Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change

Equation of the Tangent Line	Tangent Line Approximation
Equation of the Normal Line	Rates of Change and Velocity/Acceleration
Horizontal and Vertical Tangent Lines	More Practice

Note that we visited Equation of a Tangent Line here in the Definition of the Derivative section. Also, there are some Tangent Line Equation problems using the Chain Rule here in The Chain Rule section.

We will talk about the Equation of a Tangent Line with Implicit Differentiation here in the Implicit Differentiation and Related Rates section.

Equation of the Tangent Line

Let’s revisit the equation of a tangent line, which is a line that touches a curve at a point but doesn’t go through it near that point. The slope of the tangent line at this point of tangency, say “$ a$”, is the instantaneous rate of change at $ x=a$ (which we can get by taking the Derivative of the curve, as shown in the Basic Differentiation Rules section, and plugging in “$ a$” for “$ x$”). Since now we have the slope of this line, and also the coordinates of a point on the line, we can get the whole equation of this tangent line. We sometimes see this written as $ \displaystyle \frac{{dy}}{{dx}}\left| {_{{x=a}}} \right.$.

When we get the derivative of a function, we’ll use the $ x$-value of the point given to get the actual slope at that point. Then we’ll use the $ y$-value of the point to get the complete line, using either the Point-Slope $ (y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right))$ or Slope-Intercept $ (y=mx+b)$ method. I like to use point-slope. It’s really not too bad! (Weird fact: the equation of a tangent line for a linear function is just that function!)

Here are some examples. Note that in the last problem, we are given a line parallel to the tangent line, so we need to work backwards to find the point of tangency, and then find the equation of the tangent line.

Function	Equation of Tangent Line
Find the equation of the tangent line to the graph of $ \boldsymbol {f}$ at the given $ x$ value: $ \displaystyle f\left( x \right)=\frac{{2{{x}^{3}}}}{{x+1}};\,\,x=3$ Here’s what this problem looks like on a graph:	Use Quotient Rule: $ \displaystyle {f}’\left( x \right)=\frac{{\left( {x+1} \right)\left( {6{{x}^{2}}} \right)-\left( {2{{x}^{3}}} \right)\left( 1 \right)}}{{{{{\left( {x+1} \right)}}^{2}}}}=\frac{{6{{x}^{3}}+6{{x}^{2}}-2{{x}^{3}}}}{{{{{\left( {x+1} \right)}}^{2}}}}=\frac{{4{{x}^{3}}+6{{x}^{2}}}}{{{{{\left( {x+1} \right)}}^{2}}}}$ Since the derivative is a slope, we know at $ x=3$, the slope is $ \displaystyle \frac{{4{{{\left( 3 \right)}}^{3}}+6{{{\left( 3 \right)}}^{2}}}}{{{{{\left( {3+1} \right)}}^{2}}}}=\frac{{81}}{8}=10.125$. Now find the $ y$-value at $ x=3$ (use original function, since this point lies on the original function): $ \displaystyle f\left( 3 \right)=\frac{{2{{{\left( 3 \right)}}^{3}}}}{{3+1}}=\frac{{27}}{2}=13.5$. Use either the slope-intercept or point-slope method to find the equation of the line (let’s use point-slope): $ \begin{array}{c}y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,\,y-13.5=10.125\left( {x-3} \right)\\\,y-13.5=10.125x-30.375;\,\,\,y=10.125x-16.875\end{array}$ The equation of the tangent line to $ \displaystyle f\left( x \right)=\frac{{2{{x}^{3}}}}{{x+1}}$ at $ x=3$ is $ \displaystyle y=10.125x-16.875\,\,\text{or}\,\,y=\frac{{81}}{8}x-\frac{{135}}{8}$.
Find the equation of the tangent line to the graph of $ \boldsymbol {f}$ at the given point: $ \displaystyle f\left( \theta \right)=2{{\theta }^{4}}\cos \theta ;\,\,\,\left( {\frac{\pi }{2},0} \right)$	Use Product Rule: $ \displaystyle {f}’\left( x \right)=2\left[ {{{\theta }^{4}}\cdot -\sin \theta +\cos \theta \cdot 4{{\theta }^{3}}} \right]=-2{{\theta }^{4}}\sin \theta +8{{\theta }^{3}}\cos \theta $ The slope of the function is $ \displaystyle -2{{\theta }^{4}}\sin \theta +8{{\theta }^{3}}\cos \theta $, so at point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$, the slope is $ \displaystyle -2{{\left( {\frac{\pi }{2}} \right)}^{4}}\sin \frac{\pi }{2}+8{{\left( {\frac{\pi }{2}} \right)}^{3}}\cos \frac{\pi }{2}=\,-2{{\left( {\frac{\pi }{2}} \right)}^{4}}\left( 1 \right)+8{{\left( {\frac{\pi }{2}} \right)}^{3}}\left( 0 \right)=\,-\frac{{{{\pi }^{4}}}}{8}$. Use either the slope-intercept or point-slope method to find the equation of the line (let’s use slope-intercept): $ \displaystyle y=m\theta +b;\,\,\,\,\,y=-\frac{{{{\pi }^{4}}}}{8}\theta +b$. Plug in point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$ and solve for $ b$: $ \displaystyle 0=-\frac{{{{\pi }^{4}}}}{8}\left( {\frac{\pi }{2}} \right)+b;\,\,b=\frac{{{{\pi }^{5}}}}{{16}}$. The equation of the tangent line to $ f\left( \theta \right)=2{{\theta }^{4}}\cos \theta $ at the point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$ is $ \displaystyle \,\,y=-\frac{{{{\pi }^{4}}}}{8}\theta +\frac{{{{\pi }^{5}}}}{{16}}$.
Find the equation of the tangent line to the graph of $ \boldsymbol {f}$ and parallel to the give line: $ \displaystyle f\left( x \right)=\frac{1}{{\sqrt{x}}}\,\,\,(f\left( x \right)={{x}^{{-\frac{1}{2}}}})$ $ \displaystyle \text{Line:}\,\,\,\,2y+x=5$	The slope of the tangent line is $ \displaystyle m=-\frac{1}{2}$, since the tangent line is parallel to $ \displaystyle 2y+x=5,\,\text{or}\,\,y=-\frac{1}{2}x+\frac{5}{2}$ (parallel lines have same slope). The slope (derivative) of the function is $ \displaystyle {f}’\left( x \right)=-\frac{1}{2}{{x}^{{-\frac{3}{2}}}}$; solve for $ x$ to find where this tangent line “touches” the function: $ \displaystyle -\frac{1}{2}=-\frac{1}{2}{{x}^{{-\frac{3}{2}}}};\,\,\,\,{{\left( {{{x}^{{-\frac{3}{2}}}}} \right)}^{{-\frac{2}{3}}}}={{1}^{{-\frac{2}{3}}}};\,\,\,\,x\,=1$. Plug $ x=1$ into the original function to see that the point of tangency is $ (1,1)$. Use the slope-intercept method to find the equation of the line: $ \displaystyle y=-\frac{1}{2}x+b;\,\,\,\,1=-\frac{1}{2}\left( 1 \right)+b;\,\,\,\,b=1+\frac{1}{2}=\frac{3}{2}$. The equation of the function’s tangent line that is parallel to the line $ 2y+x=5$ is $ \displaystyle y=-\frac{1}{2}x+\frac{3}{2}$.
Find the equation of the tangent line(s) to the graph of f and through the given point: $ \displaystyle f\left( x \right)={{x}^{2}}\,\,\text{Point:}\,\,\left( {3,8} \right)$ Here’s a graph:	This one’s a little tricky since the given point is not on the function $ \displaystyle f\left( x \right)={{x}^{2}}$ (since $ {{3}^{2}}\ne 8$). With the tangent line in the form $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$ (point/slope), we can get the slope of $ y={{x}^{2}}$ at any point to be $ \left( {2x,y} \right)$ or $ \left( {2x,{{x}^{2}}} \right)$, since $ {y}’=2x$. Find the $ x$-values of where the tangent line touches $ y={{x}^{2}}$: $ \displaystyle \begin{align}y-8&=2x\left( {x-3} \right),\,\,\,\,y={{x}^{2}}\\{{x}^{2}}-8&=2x\left( {x-3} \right)\\{{x}^{2}}-8&=2{{x}^{2}}-6x\\{{x}^{2}}-6x+8&=0;\,\,\,\,\,\left( {x-4} \right)\left( {x-2} \right)=0;\,\,\,\,\,x=4,\,\,2\end{align}$ Thus, the tangent line touches $ y={{x}^{2}}$ at $ \left( {4,16} \right)$ and $ \left( {2,4} \right)$. The tangent lines are: $ \begin{align}y-{{y}_{1}}&=m\left( {x-{{x}_{1}}} \right)\\y-{{y}_{1}}&=\left( {2x} \right)\left( {x-{{x}_{1}}} \right)\\y-16&=2\left( 4 \right)\left( {x-4} \right)\\y&=8x-16\end{align}$ $ \begin{align}y-{{y}_{1}}&=m\left( {x-{{x}_{1}}} \right)\\y-{{y}_{1}}&=\left( {2x} \right)\left( {x-{{x}_{1}}} \right)\\y-4&=2\left( 2 \right)\left( {x-2} \right)\\y&=4x-4\end{align}$

Function

Equation of Tangent Line

Find the equation of the tangent line to the graph of $ \boldsymbol {f}$ at the given $ x$ value:

$ \displaystyle f\left( x \right)=\frac{{2{{x}^{3}}}}{{x+1}};\,\,x=3$

Here’s what this problem looks like on a graph:

Use Quotient Rule: $ \displaystyle {f}’\left( x \right)=\frac{{\left( {x+1} \right)\left( {6{{x}^{2}}} \right)-\left( {2{{x}^{3}}} \right)\left( 1 \right)}}{{{{{\left( {x+1} \right)}}^{2}}}}=\frac{{6{{x}^{3}}+6{{x}^{2}}-2{{x}^{3}}}}{{{{{\left( {x+1} \right)}}^{2}}}}=\frac{{4{{x}^{3}}+6{{x}^{2}}}}{{{{{\left( {x+1} \right)}}^{2}}}}$

Since the derivative is a slope, we know at $ x=3$, the slope is $ \displaystyle \frac{{4{{{\left( 3 \right)}}^{3}}+6{{{\left( 3 \right)}}^{2}}}}{{{{{\left( {3+1} \right)}}^{2}}}}=\frac{{81}}{8}=10.125$. Now find the $ y$-value at $ x=3$ (use original function, since this point lies on the original function): $ \displaystyle f\left( 3 \right)=\frac{{2{{{\left( 3 \right)}}^{3}}}}{{3+1}}=\frac{{27}}{2}=13.5$.

Use either the slope-intercept or point-slope method to find the equation of the line (let’s use point-slope):

$ \begin{array}{c}y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,\,y-13.5=10.125\left( {x-3} \right)\\\,y-13.5=10.125x-30.375;\,\,\,y=10.125x-16.875\end{array}$

The equation of the tangent line to $ \displaystyle f\left( x \right)=\frac{{2{{x}^{3}}}}{{x+1}}$ at $ x=3$ is $ \displaystyle y=10.125x-16.875\,\,\text{or}\,\,y=\frac{{81}}{8}x-\frac{{135}}{8}$.

Find the equation of the tangent line to the graph of $ \boldsymbol {f}$ at the given point:

$ \displaystyle f\left( \theta \right)=2{{\theta }^{4}}\cos \theta ;\,\,\,\left( {\frac{\pi }{2},0} \right)$

Use Product Rule: $ \displaystyle {f}’\left( x \right)=2\left[ {{{\theta }^{4}}\cdot -\sin \theta +\cos \theta \cdot 4{{\theta }^{3}}} \right]=-2{{\theta }^{4}}\sin \theta +8{{\theta }^{3}}\cos \theta $

The slope of the function is $ \displaystyle -2{{\theta }^{4}}\sin \theta +8{{\theta }^{3}}\cos \theta $, so at point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$, the slope is $ \displaystyle -2{{\left( {\frac{\pi }{2}} \right)}^{4}}\sin \frac{\pi }{2}+8{{\left( {\frac{\pi }{2}} \right)}^{3}}\cos \frac{\pi }{2}=\,-2{{\left( {\frac{\pi }{2}} \right)}^{4}}\left( 1 \right)+8{{\left( {\frac{\pi }{2}} \right)}^{3}}\left( 0 \right)=\,-\frac{{{{\pi }^{4}}}}{8}$.

Use either the slope-intercept or point-slope method to find the equation of the line (let’s use slope-intercept): $ \displaystyle y=m\theta +b;\,\,\,\,\,y=-\frac{{{{\pi }^{4}}}}{8}\theta +b$. Plug in point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$ and solve for $ b$: $ \displaystyle 0=-\frac{{{{\pi }^{4}}}}{8}\left( {\frac{\pi }{2}} \right)+b;\,\,b=\frac{{{{\pi }^{5}}}}{{16}}$.

The equation of the tangent line to $ f\left( \theta \right)=2{{\theta }^{4}}\cos \theta $ at the point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$ is $ \displaystyle \,\,y=-\frac{{{{\pi }^{4}}}}{8}\theta +\frac{{{{\pi }^{5}}}}{{16}}$.

Find the equation of the tangent line to the graph of $ \boldsymbol {f}$ and parallel to the give line:

$ \displaystyle f\left( x \right)=\frac{1}{{\sqrt{x}}}\,\,\,(f\left( x \right)={{x}^{{-\frac{1}{2}}}})$

$ \displaystyle \text{Line:}\,\,\,\,2y+x=5$

The slope of the tangent line is $ \displaystyle m=-\frac{1}{2}$, since the tangent line is parallel to $ \displaystyle 2y+x=5,\,\text{or}\,\,y=-\frac{1}{2}x+\frac{5}{2}$ (parallel lines have same slope). The slope (derivative) of the function is $ \displaystyle {f}’\left( x \right)=-\frac{1}{2}{{x}^{{-\frac{3}{2}}}}$; solve for $ x$ to find where this tangent line “touches” the function: $ \displaystyle -\frac{1}{2}=-\frac{1}{2}{{x}^{{-\frac{3}{2}}}};\,\,\,\,{{\left( {{{x}^{{-\frac{3}{2}}}}} \right)}^{{-\frac{2}{3}}}}={{1}^{{-\frac{2}{3}}}};\,\,\,\,x\,=1$.

Plug $ x=1$ into the original function to see that the point of tangency is $ (1,1)$. Use the slope-intercept method to find the equation of the line: $ \displaystyle y=-\frac{1}{2}x+b;\,\,\,\,1=-\frac{1}{2}\left( 1 \right)+b;\,\,\,\,b=1+\frac{1}{2}=\frac{3}{2}$.

The equation of the function’s tangent line that is parallel to the line $ 2y+x=5$ is $ \displaystyle y=-\frac{1}{2}x+\frac{3}{2}$.

Find the equation of the tangent line(s) to the graph of f and through the given point:

$ \displaystyle f\left( x \right)={{x}^{2}}\,\,\text{Point:}\,\,\left( {3,8} \right)$

Here’s a graph:

This one’s a little tricky since the given point is not on the function $ \displaystyle f\left( x \right)={{x}^{2}}$ (since $ {{3}^{2}}\ne 8$).

With the tangent line in the form $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$ (point/slope), we can get the slope of $ y={{x}^{2}}$ at any point to be $ \left( {2x,y} \right)$ or $ \left( {2x,{{x}^{2}}} \right)$, since $ {y}’=2x$. Find the $ x$-values of where the tangent line touches $ y={{x}^{2}}$:

$ \displaystyle \begin{align}y-8&=2x\left( {x-3} \right),\,\,\,\,y={{x}^{2}}\\{{x}^{2}}-8&=2x\left( {x-3} \right)\\{{x}^{2}}-8&=2{{x}^{2}}-6x\\{{x}^{2}}-6x+8&=0;\,\,\,\,\,\left( {x-4} \right)\left( {x-2} \right)=0;\,\,\,\,\,x=4,\,\,2\end{align}$

Thus, the tangent line touches $ y={{x}^{2}}$ at $ \left( {4,16} \right)$ and $ \left( {2,4} \right)$. The tangent lines are:

$ \begin{align}y-{{y}_{1}}&=m\left( {x-{{x}_{1}}} \right)\\y-{{y}_{1}}&=\left( {2x} \right)\left( {x-{{x}_{1}}} \right)\\y-16&=2\left( 4 \right)\left( {x-4} \right)\\y&=8x-16\end{align}$ $ \begin{align}y-{{y}_{1}}&=m\left( {x-{{x}_{1}}} \right)\\y-{{y}_{1}}&=\left( {2x} \right)\left( {x-{{x}_{1}}} \right)\\y-4&=2\left( 2 \right)\left( {x-2} \right)\\y&=4x-4\end{align}$

Here are other types of problems you might see:

Slope Derivative Problem	Solution
The tangent line to the graph of $ f\left( x \right)$ at the point $ \left( {2,1} \right)$ goes through the point $ \left( {-1,4} \right)$. Find $ {f}’\left(2\right)$.	This seems complicated, but we just want the slope of the line that is tangent to the point $ \left( {2,1} \right)$ at this point ($ x=2$). Since this tangent line also goes through the point $ \left( {-1,4} \right)$, we simply need the slope of the line between the two points: $ \displaystyle m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}=\frac{{4-1}}{{-1-2}}=-1$. Thus, $ {f}’\left( 2 \right)=-1$.
For what point on the curve of $ y=2{{x}^{2}}-2x$ does the slope equal $ 10$?	To get the slope of the function at a certain point, take the derivative: $ {y}’=4x-2$. Then set this to $ 10$ (the slope given) and solve for $ x$: $ 4x-2=10;\,\,\,x=3$. To get the $ y$-value of the point, plug $ 3$ into the original equation: $ y=2{{\left( 3 \right)}^{2}}-2\left( 3 \right)=12$. Thus, the point on the curve $ y=2{{x}^{2}}-2x$ where the slope is $ 10$ is $ \left( {3,12} \right)$. Not too bad!

Slope Derivative Problem

Solution

The tangent line to the graph of $ f\left( x \right)$ at the point $ \left( {2,1} \right)$ goes through the point $ \left( {-1,4} \right)$.

Find $ {f}’\left(2\right)$.

This seems complicated, but we just want the slope of the line that is tangent to the point $ \left( {2,1} \right)$ at this point ($ x=2$). Since this tangent line also goes through the point $ \left( {-1,4} \right)$, we simply need the slope of the line between the two points: $ \displaystyle m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}=\frac{{4-1}}{{-1-2}}=-1$.

Thus, $ {f}’\left( 2 \right)=-1$.

For what point on the curve of $ y=2{{x}^{2}}-2x$ does the slope equal $ 10$?

To get the slope of the function at a certain point, take the derivative: $ {y}’=4x-2$. Then set this to $ 10$ (the slope given) and solve for $ x$: $ 4x-2=10;\,\,\,x=3$. To get the $ y$-value of the point, plug $ 3$ into the original equation: $ y=2{{\left( 3 \right)}^{2}}-2\left( 3 \right)=12$.

Thus, the point on the curve $ y=2{{x}^{2}}-2x$ where the slope is $ 10$ is $ \left( {3,12} \right)$. Not too bad!

Here is one more; in this problem, we’ll find the equation of a parabola that goes through a certain point and is tangent to a line at another point.

Tangent Line Parabola Problem	Solution
The graph of the parabola $ y=a{{x}^{2}}+bx+c$ goes through the point $ \left( {0,1} \right)$, and is tangent to the line $ y=4x-2$ at the point $ \left( {1,2} \right)$. Find the equation of this parabola.	Typically, the trick to doing problems like this is to try to come up with a System of Equations with the same number of variables as equations. Since the equation of a parabola is $ y=a{{x}^{2}}+bx+c$, and the parabola goes through the point $ \left( {0,1} \right)$ (easiest point – this is the $ y$-intercept), we have $ 1=a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c$, so $ c=1$. The equation of the parabola so far is $ y=a{{x}^{2}}+bx+1$. Since the parabola is tangent to the line $ y=4x-2$ at point $ \left( {1,2} \right)$, the slope (derivative) of the parabola is $ 4$ at this point. Take the derivative of the parabola function to get its slope (don’t let the “letters” scare you): $ \begin{array}{l}y=a{{x}^{2}}+bx+c\\{y}’=2ax+b\end{array}$ Now we know that $ 2ax+b=4$, so at point $ \left( {1,2} \right)$, plug in $ 1$ for $ x$ to get $ 2a\left( 1 \right)+b=4$, or $ 2a+b=4$. Since the point of tangency at $ \left( {1,2} \right)$ is also on the parabola, we can plug this point into $ y=a{{x}^{2}}+bx+1$ to get $ 2=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+1$, or $ a+b=1$. We have a system with two equations and two unknowns; solve for $ a$ and $ b$: $ \begin{align}2a+b&=4\\a+b&=1\end{align}$ $ \displaystyle \begin{array}{l}\underline{\begin{array}{l}\,\,\,\,2a\,+\,b=4\\-a+-b=-1\end{array}}\\\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,=3;\,\,\,\,\,b=-2\end{array}$ The equation of the parabola is $ y=3{{x}^{2}}-2x+1$.

Tangent Line Parabola Problem

Solution

The graph of the parabola $ y=a{{x}^{2}}+bx+c$ goes through the point $ \left( {0,1} \right)$, and is tangent to the line $ y=4x-2$ at the point $ \left( {1,2} \right)$.

Find the equation of this parabola.

Typically, the trick to doing problems like this is to try to come up with a System of Equations with the same number of variables as equations.

Since the equation of a parabola is $ y=a{{x}^{2}}+bx+c$, and the parabola goes through the point $ \left( {0,1} \right)$ (easiest point – this is the $ y$-intercept), we have $ 1=a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c$, so $ c=1$. The equation of the parabola so far is $ y=a{{x}^{2}}+bx+1$.

Since the parabola is tangent to the line $ y=4x-2$ at point $ \left( {1,2} \right)$, the slope (derivative) of the parabola is $ 4$ at this point. Take the derivative of the parabola function to get its slope (don’t let the “letters” scare you):

$ \begin{array}{l}y=a{{x}^{2}}+bx+c\\{y}’=2ax+b\end{array}$

Now we know that $ 2ax+b=4$, so at point $ \left( {1,2} \right)$, plug in $ 1$ for $ x$ to get $ 2a\left( 1 \right)+b=4$, or $ 2a+b=4$. Since the point of tangency at $ \left( {1,2} \right)$ is also on the parabola, we can plug this point into $ y=a{{x}^{2}}+bx+1$ to get $ 2=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+1$, or $ a+b=1$.

We have a system with two equations and two unknowns; solve for $ a$ and $ b$:

$ \begin{align}2a+b&=4\\a+b&=1\end{align}$ $ \displaystyle \begin{array}{l}\underline{\begin{array}{l}\,\,\,\,2a\,+\,b=4\\-a+-b=-1\end{array}}\\\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,=3;\,\,\,\,\,b=-2\end{array}$

The equation of the parabola is $ y=3{{x}^{2}}-2x+1$.

Equation of the Normal Line

The normal line to a curve at a point is the line through that point that is perpendicular to the tangent. Remember that a line is perpendicular to another line if their slopes are opposite reciprocals of each other; for example, if one slope is $ 4$, the other slope would be $ \displaystyle -\frac{1}{4}$.

We do this problem the same way, but use the opposite reciprocal of the slope when we go to get the line; here is an example:

Function and Graph	Equation of the Normal Line
Find the equation of the normal line to the graph of $ f$ at the given $ x$-value: $ \displaystyle f\left( x \right)=\frac{2}{{\sqrt{{x+1}}}};\,\,\,x=8$	Use Power Rule: $ \begin{align}f\left( x \right)&=2{{\left( {x+1} \right)}^{{-\frac{1}{2}}}}\\{f}’\left( x \right)&=2\cdot -\frac{1}{2}{{\left( {x+1} \right)}^{{-\frac{3}{2}}}}=-{{\left( {x+1} \right)}^{{-\frac{3}{2}}}}=-\frac{1}{{{{{\left( {x+1} \right)}}^{{\frac{3}{2}}}}}}\end{align}$ Since the derivative is a slope, at $ x=8$, the slope is $ \displaystyle -\frac{1}{{{{{\left( {8+1} \right)}}^{{\frac{3}{2}}}}}}=-\frac{1}{{27}}$. The slope of the line normal to this line is the opposite reciprocal, which is $ 27$. Now we need to find the $ y$-value at $ x=8$ (use the original function, since this point lies on the original function): $ \displaystyle f\left( 8 \right)=\frac{2}{{\sqrt{{8+1}}}}=\frac{2}{3}$. We can use either the slope-intercept or point-slope method to find the equation of the normal line (let’s use point-slope): $ \displaystyle y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,\,\,y-\frac{2}{3}=27\left( {x-8} \right);\,\,\,\,y-\frac{2}{3}=27x-216;\,\,\,\,\,y=27x-215\frac{1}{3}$. The equation of the normal line to $ \displaystyle f\left( x \right)=\frac{2}{{\sqrt{{x+1}}}}$ at $ x=8$ is $ \displaystyle y=27x-215\frac{1}{3}\,\,\text{or}\,\,y=27x-\frac{{646}}{3}$.

Function and Graph

Equation of the Normal Line

Find the equation of the normal line to the graph of $ f$ at the given $ x$-value:

$ \displaystyle f\left( x \right)=\frac{2}{{\sqrt{{x+1}}}};\,\,\,x=8$

Use Power Rule: $ \begin{align}f\left( x \right)&=2{{\left( {x+1} \right)}^{{-\frac{1}{2}}}}\\{f}’\left( x \right)&=2\cdot -\frac{1}{2}{{\left( {x+1} \right)}^{{-\frac{3}{2}}}}=-{{\left( {x+1} \right)}^{{-\frac{3}{2}}}}=-\frac{1}{{{{{\left( {x+1} \right)}}^{{\frac{3}{2}}}}}}\end{align}$

Since the derivative is a slope, at $ x=8$, the slope is $ \displaystyle -\frac{1}{{{{{\left( {8+1} \right)}}^{{\frac{3}{2}}}}}}=-\frac{1}{{27}}$. The slope of the line normal to this line is the opposite reciprocal, which is $ 27$. Now we need to find the $ y$-value at $ x=8$ (use the original function, since this point lies on the original function): $ \displaystyle f\left( 8 \right)=\frac{2}{{\sqrt{{8+1}}}}=\frac{2}{3}$.

We can use either the slope-intercept or point-slope method to find the equation of the normal line (let’s use point-slope): $ \displaystyle y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,\,\,y-\frac{2}{3}=27\left( {x-8} \right);\,\,\,\,y-\frac{2}{3}=27x-216;\,\,\,\,\,y=27x-215\frac{1}{3}$.

The equation of the normal line to $ \displaystyle f\left( x \right)=\frac{2}{{\sqrt{{x+1}}}}$ at $ x=8$ is $ \displaystyle y=27x-215\frac{1}{3}\,\,\text{or}\,\,y=27x-\frac{{646}}{3}$.

Horizontal and Vertical Tangent Lines

Sometimes we want to know at what point(s) a function has either a horizontal or vertical tangent line (if they exist). For a horizontal tangent line ($ 0$ slope), we want to get the derivative, and set it to $ 0$, which means setting the numerator to $ \boldsymbol {0}$. Then we can get the $ x$-value, and use the original function to get the $ y$-value; thus, we have the point(s) at the horizontal tangent line. We also have to make sure that that the denominator isn’t $ 0$ at these points.

For a vertical tangent line (undefined slope), we want to get the derivative, and set the bottom or denominator to $ \boldsymbol {0}$. Then we can get the $ x$-value, and use the original function to get the $ y$-value; thus, we have the point(s) at the vertical tangent line. We also to have to make sure the numerator isn’t $ 0$ at these points. Note that where functions have vertical tangent lines, they are not differentiable at that point.

Here are some examples:

Horizontal and Vertical Tangent Problem	Solution
Find the point(s) for which $ \displaystyle f\left( x \right)=x+\frac{5}{x}$ has a horizontal tangent line and a vertical tangent line. Graph of function:	First take the derivative of the function $ f\left( x \right)=x+5{{x}^{{-1}}}$, which is $ \displaystyle {f}’\left( x \right)=1-5{{x}^{{-2}}}=1-\frac{5}{{{{x}^{2}}}}=\frac{{{{x}^{2}}-5}}{{{{x}^{2}}}}$. To get the point(s) of the horizontal tangent line, set this to $ 0$ (which means setting the numerator to $ 0$) and solve for $ x$: $ \displaystyle \frac{{{{x}^{2}}-5}}{{{{x}^{2}}}}=0;\,\,\,\,{{x}^{2}}-5=0;\,\,\,\,\,x=\pm \sqrt{5}$. Plug these $ x$-values into the original function to get the corresponding $ y$-values: $ \displaystyle f\left( {\sqrt{5}} \right)=\sqrt{5}+\frac{5}{{\sqrt{5}}}\cdot \frac{{\sqrt{5}}}{{\sqrt{5}}}=2\sqrt{5}$. Similarly, $ \displaystyle f\left( {-\sqrt{5}} \right)=-\sqrt{5}+\frac{5}{{-\sqrt{5}}}\cdot \frac{{\sqrt{5}}}{{\sqrt{5}}}=-2\sqrt{5}$. Thus, the points where there is a horizontal tangent line are $ \left( {\sqrt{5},2\sqrt{5}} \right)$ and $ \left( {-\sqrt{5},-2\sqrt{5}} \right)$. To get the point(s) of the vertical tangent line, set the denominator to $ 0$ and solve for $ x$: $ \displaystyle {{x}^{2}}=0;\,\,\,\,x=0$. When we plug this $ x$-value into the original function to get the corresponding $ y$-value, we see that the function is undefined (because of dividing by $ 0$), so there are no vertical tangent lines. (Note that the function is non-differentiable at $ x=0$, where there’s a vertical asymptote).
Find the point(s) for which $ \displaystyle f\left( x \right)=\sqrt[3]{{x-1}}$ has a horizontal tangent line and a vertical tangent line. Graph of function:	First take the derivative of the function $ \displaystyle f\left( x \right)={{\left( {x-1} \right)}^{{\frac{1}{3}}}}$, which is $ \displaystyle {f}’\left( x \right)=\frac{1}{3}{{\left( {x-1} \right)}^{{-\frac{2}{3}}}}=\frac{1}{{3\sqrt[3]{{{{{\left( {x-1} \right)}}^{2}}}}}}$. We can see that the numerator can never be $ 0$, so there are no horizontal tangent lines. To get the point(s) of the vertical tangent line, set the denominator to $ 0$ and solve for $ x$: $ \displaystyle 3\sqrt[3]{{{{{\left( {x-1} \right)}}^{2}}}}=0;\,\,{{\left( {x-1} \right)}^{2}}=0;\,\,\,x=1$. Plug this $ x$-value into the original function to get the corresponding $ y$-value: $ f\left( 0 \right)=\sqrt[3]{{1-1}}=0$; the point where there is a vertical tangent line is $ \left( {1,0} \right)$. Note that this point is non-differentiable.
Find the point(s) for which $ f\left( x \right)={{x}^{{\frac{3}{2}}}}-{{x}^{{\frac{1}{2}}}}$ has a horizontal tangent line and a vertical tangent line. Graph of function:	Take the derivative of the function $ f\left( x \right)={{x}^{{\frac{3}{2}}}}-{{x}^{{\frac{1}{2}}}}$, which is $ \displaystyle {f}’\left( x \right)=\frac{3}{2}{{x}^{{\frac{1}{2}}}}-\frac{1}{2}{{x}^{{-\frac{1}{2}}}}=\frac{{3\sqrt{x}}}{2}-\frac{1}{{2\sqrt{x}}}=\frac{{3\sqrt{x}}}{2}\cdot \frac{{\sqrt{x}}}{{\sqrt{x}}}-\frac{1}{{2\sqrt{x}}}=\frac{{3x-1}}{{2\sqrt{x}}}$. To get the point(s) of the horizontal tangent line, set this to $ 0$ (which means setting the numerator to $ 0$), and solve for $ x$: $ \displaystyle \frac{{3x-1}}{{2\sqrt{x}}}=0;\,\,3x-1=0;\,\,\,x=\frac{1}{3}$. (We notice that this $ x$-value doesn’t make the denominator $ 0$). Plug this $ x$-value into the original function to get the corresponding $ y$ value: $ \displaystyle f\left( {\frac{1}{3}} \right)={{\left( {\frac{1}{3}} \right)}^{{\frac{3}{2}}}}-\,\,{{\left( {\frac{1}{3}} \right)}^{{\frac{1}{2}}}}=-\frac{{2\sqrt{3}}}{9}$. The point where there is a horizontal tangent line is $ \displaystyle \left( {\frac{1}{3},-\frac{{2\sqrt{3}}}{9}} \right)$. To get the point(s) of the vertical tangent line, set the denominator to $ 0$ and solve for $ x$: $ \displaystyle 2\sqrt{x}=0;\,\,x=0$ (We notice that this $ x$-value doesn’t make the numerator $ 0$). Plug this $ x$-value into the original function to get the corresponding $ y$-value: $ \displaystyle f\left( 0 \right)={{0}^{{\frac{3}{2}}}}-{{0}^{{\frac{1}{2}}}}=0$; the point where there is a vertical tangent line is $ \left( {0,0} \right)$ (on an endpoint!). Again, note that this point is-non-differentiable.

Horizontal and Vertical Tangent Problem

Solution

Find the point(s) for which $ \displaystyle f\left( x \right)=x+\frac{5}{x}$ has a horizontal tangent line and a vertical tangent line.

Graph of function:

First take the derivative of the function $ f\left( x \right)=x+5{{x}^{{-1}}}$, which is $ \displaystyle {f}’\left( x \right)=1-5{{x}^{{-2}}}=1-\frac{5}{{{{x}^{2}}}}=\frac{{{{x}^{2}}-5}}{{{{x}^{2}}}}$. To get the point(s) of the horizontal tangent line, set this to $ 0$ (which means setting the numerator to $ 0$) and solve for $ x$: $ \displaystyle \frac{{{{x}^{2}}-5}}{{{{x}^{2}}}}=0;\,\,\,\,{{x}^{2}}-5=0;\,\,\,\,\,x=\pm \sqrt{5}$.

Plug these $ x$-values into the original function to get the corresponding $ y$-values: $ \displaystyle f\left( {\sqrt{5}} \right)=\sqrt{5}+\frac{5}{{\sqrt{5}}}\cdot \frac{{\sqrt{5}}}{{\sqrt{5}}}=2\sqrt{5}$. Similarly, $ \displaystyle f\left( {-\sqrt{5}} \right)=-\sqrt{5}+\frac{5}{{-\sqrt{5}}}\cdot \frac{{\sqrt{5}}}{{\sqrt{5}}}=-2\sqrt{5}$. Thus, the points where there is a horizontal tangent line are $ \left( {\sqrt{5},2\sqrt{5}} \right)$ and $ \left( {-\sqrt{5},-2\sqrt{5}} \right)$.

To get the point(s) of the vertical tangent line, set the denominator to $ 0$ and solve for $ x$: $ \displaystyle {{x}^{2}}=0;\,\,\,\,x=0$. When we plug this $ x$-value into the original function to get the corresponding $ y$-value, we see that the function is undefined (because of dividing by $ 0$), so there are no vertical tangent lines. (Note that the function is non-differentiable at $ x=0$, where there’s a vertical asymptote).

Find the point(s) for which $ \displaystyle f\left( x \right)=\sqrt[3]{{x-1}}$ has a horizontal tangent line and a vertical tangent line.

Graph of function:

First take the derivative of the function $ \displaystyle f\left( x \right)={{\left( {x-1} \right)}^{{\frac{1}{3}}}}$, which is $ \displaystyle {f}’\left( x \right)=\frac{1}{3}{{\left( {x-1} \right)}^{{-\frac{2}{3}}}}=\frac{1}{{3\sqrt[3]{{{{{\left( {x-1} \right)}}^{2}}}}}}$. We can see that the numerator can never be $ 0$, so there are no horizontal tangent lines.

To get the point(s) of the vertical tangent line, set the denominator to $ 0$ and solve for $ x$: $ \displaystyle 3\sqrt[3]{{{{{\left( {x-1} \right)}}^{2}}}}=0;\,\,{{\left( {x-1} \right)}^{2}}=0;\,\,\,x=1$.

Plug this $ x$-value into the original function to get the corresponding $ y$-value: $ f\left( 0 \right)=\sqrt[3]{{1-1}}=0$; the point where there is a vertical tangent line is $ \left( {1,0} \right)$. Note that this point is non-differentiable.

Find the point(s) for which $ f\left( x \right)={{x}^{{\frac{3}{2}}}}-{{x}^{{\frac{1}{2}}}}$ has a horizontal tangent line and a vertical tangent line.

Graph of function:

Take the derivative of the function $ f\left( x \right)={{x}^{{\frac{3}{2}}}}-{{x}^{{\frac{1}{2}}}}$, which is $ \displaystyle {f}’\left( x \right)=\frac{3}{2}{{x}^{{\frac{1}{2}}}}-\frac{1}{2}{{x}^{{-\frac{1}{2}}}}=\frac{{3\sqrt{x}}}{2}-\frac{1}{{2\sqrt{x}}}=\frac{{3\sqrt{x}}}{2}\cdot \frac{{\sqrt{x}}}{{\sqrt{x}}}-\frac{1}{{2\sqrt{x}}}=\frac{{3x-1}}{{2\sqrt{x}}}$. To get the point(s) of the horizontal tangent line, set this to $ 0$ (which means setting the numerator to $ 0$), and solve for $ x$: $ \displaystyle \frac{{3x-1}}{{2\sqrt{x}}}=0;\,\,3x-1=0;\,\,\,x=\frac{1}{3}$. (We notice that this $ x$-value doesn’t make the denominator $ 0$).

Plug this $ x$-value into the original function to get the corresponding $ y$ value: $ \displaystyle f\left( {\frac{1}{3}} \right)={{\left( {\frac{1}{3}} \right)}^{{\frac{3}{2}}}}-\,\,{{\left( {\frac{1}{3}} \right)}^{{\frac{1}{2}}}}=-\frac{{2\sqrt{3}}}{9}$. The point where there is a horizontal tangent line is $ \displaystyle \left( {\frac{1}{3},-\frac{{2\sqrt{3}}}{9}} \right)$.

To get the point(s) of the vertical tangent line, set the denominator to $ 0$ and solve for $ x$: $ \displaystyle 2\sqrt{x}=0;\,\,x=0$ (We notice that this $ x$-value doesn’t make the numerator $ 0$). Plug this $ x$-value into the original function to get the corresponding $ y$-value: $ \displaystyle f\left( 0 \right)={{0}^{{\frac{3}{2}}}}-{{0}^{{\frac{1}{2}}}}=0$; the point where there is a vertical tangent line is $ \left( {0,0} \right)$ (on an endpoint!). Again, note that this point is-non-differentiable.

Tangent Line Approximation

Also known as local linearization and linear approximation, it turns out that we can use a tangent line to approximate the value or graph of a function. The reason they want you to learn this stuff is so you can “appreciate the math” and how they used to use Calculus many years ago before fancy calculators and computers. And, actually, it’s pretty neat! (Note that we will show linear approximation using differentials here in the Differentials, Linear Approximation and Error Propagation section.)

Let’s say we have a function, such as $ f\left( x \right)=\sqrt{x}$, and we want to know what the value of this function is ($ f\left( x \right)$), when $ x=4.023$. We know it’s close to $ 2$, but we can get even closer without a calculator using Calculus. We can do this by using the point-slope formula $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, where $ \left( {{{x}_{1}},\,\,{{y}_{1}}} \right)$ is a point that we can get easily, like $ \left( {4,2} \right)$, and $ m$ is the derivative of the function at that point. We can then get our new $ y$-value, by plugging the original $ x$-value we have ($ 4.03$).

Here is this problem, along with some others; use linear approximation (local linearization, or tangent line approximation) to find an approximation for the indicated values:

Problem and Graph	Linear Approximation
$ \displaystyle \begin{array}{c}f\left( x \right)=\sqrt{x}\\\text{Find }f\left( {4.03} \right).\end{array}$	The trick is to find the tangent line approximation at a point that is simple and obvious; this point is $ \left( {4,2} \right)$, since $ 2=\sqrt{4}$, and $ 4$ is close to $ 4.03$. To get the slope of the tangent line for the point-slope equation $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, take the derivative of the function: $ \displaystyle f\left( x \right)=\sqrt{x};\,\,f\left( x \right)={{x}^{{\frac{1}{2}}}};\,\,{f}’\left( x \right)=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$. Then plug in $ x=4$ to get the slope at the point $ \left( {4,2} \right)$: $ \displaystyle {f}’\left( 4 \right)=\frac{1}{2}{{\left( 4 \right)}^{{-\frac{1}{2}}}}=\frac{1}{2}\left( {\frac{1}{{\sqrt{4}}}} \right)=\frac{1}{4}$. Use the tangent line equation to get the “new” $ y$ by plugging in the slope and the point $ \left( {4,2} \right)$: $ \displaystyle y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,y-2=\left( {\frac{1}{4}} \right)\left( {x-4} \right);\,\,y=\frac{1}{4}x+1$. Plug in the $ x$ we want from the problem, which is $ 4.03$: $ \displaystyle y=\frac{1}{4}x+1;\,\,y=\frac{1}{4}\left( {4.03} \right)+1;\,\,\,y=2.0075$. Pretty amazing, since $ \sqrt{{4.03}}\approx 2.007486$.
$ \begin{array}{c}f\left( x \right)={{x}^{3}}-4x+3\\\text{A point on }f\text{ is }\left( {2,3} \right);\,\,\text{find }f\left( {2.05} \right).\end{array}$	In this problem, we’re given the point of the tangent $ \left( {2,3} \right)$, so we just need to get the slope of the function at that point. To get the slope of the tangent line for the point-slope equation $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, take the derivative of the function $ f\left( x \right)={{x}^{3}}-4x+3;\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}-4$. Plug in $ 2$ to get the slope at the point $ \left( {2,3} \right)$: $ {f}’\left( 2 \right)=3{{\left( 2 \right)}^{2}}-4=8$. Now use the tangent line equation to get the “new” $ y$ by plugging in the slope and the point $ \left( {2,3} \right)$: $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,y-3=\left( 8 \right)\left( {x-2} \right);\,\,\,\,y=8x-13$. Plug in the $ x$ we want from the problem, which is $ 2.05$: $ y=8x-13;\,\,y=8\left( {2.05} \right)-13;\,\,y=3.4$. Pretty cool, since $ {{\left( {2.05} \right)}^{3}}-4\left( {2.05} \right)+3=3.415125$.
$ \begin{array}{c}f\left( x \right)=\sin \left( x \right)\\\text{find }f\left( {3.16} \right)\end{array}$	We can find a tangent line approximation at a point that is simple and obvious; this point is $ \left( {\pi ,0} \right)$. (Use $ \pi $, since it is close to $ 3.16$). To get the slope of the tangent line for the point-slope equation $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, take the derivative of the function: $ f\left( x \right)=\sin \left( x \right);\,\,{f}’\left( x \right)=\cos \left( x \right)$. Plug in $ \pi $ to get the slope at the point $ {f}’\left( \pi \right)=\cos \left( \pi \right)=-1$. Use the tangent line equation to get the “new” $ y$ by plugging in the slope and the point $ \left( {\pi ,0} \right)$: $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,\,y-0=\left( {-1} \right)\left( {x-\pi } \right);\,\,\,\,y=-x+\pi $. Plug in the $ x$ we want from the problem, which is $ 3.16$: $ y=-x+\pi ;\,\,y=-3.16+\pi ;\,\,y\approx -.018407$. Pretty nice, since $ \sin \left( {3.16} \right)\approx -.018406$.

Problem and Graph

Linear Approximation

$ \displaystyle \begin{array}{c}f\left( x \right)=\sqrt{x}\\\text{Find }f\left( {4.03} \right).\end{array}$

The trick is to find the tangent line approximation at a point that is simple and obvious; this point is $ \left( {4,2} \right)$, since $ 2=\sqrt{4}$, and $ 4$ is close to $ 4.03$.

To get the slope of the tangent line for the point-slope equation $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, take the derivative of the function: $ \displaystyle f\left( x \right)=\sqrt{x};\,\,f\left( x \right)={{x}^{{\frac{1}{2}}}};\,\,{f}’\left( x \right)=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$. Then plug in $ x=4$ to get the slope at the point $ \left( {4,2} \right)$: $ \displaystyle {f}’\left( 4 \right)=\frac{1}{2}{{\left( 4 \right)}^{{-\frac{1}{2}}}}=\frac{1}{2}\left( {\frac{1}{{\sqrt{4}}}} \right)=\frac{1}{4}$.

Use the tangent line equation to get the “new” $ y$ by plugging in the slope and the point $ \left( {4,2} \right)$: $ \displaystyle y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,y-2=\left( {\frac{1}{4}} \right)\left( {x-4} \right);\,\,y=\frac{1}{4}x+1$.

Plug in the $ x$ we want from the problem, which is $ 4.03$: $ \displaystyle y=\frac{1}{4}x+1;\,\,y=\frac{1}{4}\left( {4.03} \right)+1;\,\,\,y=2.0075$. Pretty amazing, since $ \sqrt{{4.03}}\approx 2.007486$.

$ \begin{array}{c}f\left( x \right)={{x}^{3}}-4x+3\\\text{A point on }f\text{ is }\left( {2,3} \right);\,\,\text{find }f\left( {2.05} \right).\end{array}$

In this problem, we’re given the point of the tangent $ \left( {2,3} \right)$, so we just need to get the slope of the function at that point.

To get the slope of the tangent line for the point-slope equation $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, take the derivative of the function $ f\left( x \right)={{x}^{3}}-4x+3;\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}-4$. Plug in $ 2$ to get the slope at the point $ \left( {2,3} \right)$: $ {f}’\left( 2 \right)=3{{\left( 2 \right)}^{2}}-4=8$.

Now use the tangent line equation to get the “new” $ y$ by plugging in the slope and the point $ \left( {2,3} \right)$: $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,y-3=\left( 8 \right)\left( {x-2} \right);\,\,\,\,y=8x-13$.

Plug in the $ x$ we want from the problem, which is $ 2.05$: $ y=8x-13;\,\,y=8\left( {2.05} \right)-13;\,\,y=3.4$. Pretty cool, since $ {{\left( {2.05} \right)}^{3}}-4\left( {2.05} \right)+3=3.415125$.

$ \begin{array}{c}f\left( x \right)=\sin \left( x \right)\\\text{find }f\left( {3.16} \right)\end{array}$

We can find a tangent line approximation at a point that is simple and obvious; this point is $ \left( {\pi ,0} \right)$. (Use $ \pi $, since it is close to $ 3.16$).

To get the slope of the tangent line for the point-slope equation $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, take the derivative of the function: $ f\left( x \right)=\sin \left( x \right);\,\,{f}’\left( x \right)=\cos \left( x \right)$. Plug in $ \pi $ to get the slope at the point $ {f}’\left( \pi \right)=\cos \left( \pi \right)=-1$.

Use the tangent line equation to get the “new” $ y$ by plugging in the slope and the point $ \left( {\pi ,0} \right)$: $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,\,y-0=\left( {-1} \right)\left( {x-\pi } \right);\,\,\,\,y=-x+\pi $.

Plug in the $ x$ we want from the problem, which is $ 3.16$: $ y=-x+\pi ;\,\,y=-3.16+\pi ;\,\,y\approx -.018407$. Pretty nice, since $ \sin \left( {3.16} \right)\approx -.018406$.

Rates of Change and Velocity/Acceleration

Note that there are more position, velocity, and acceleration problems here in the Antiderivatives and Indefinite Integration section. and in the Integration as Accumulated Change section.

The derivative has many applications in “real life”; one of the most useful is to find the rate of change of one variable with respect to another. Think of a rate of change, or sometimes called an instantaneous rate of change as how fast something is changing at a certain point, like a point in time. For example, if a particle is moving along a horizontal line, it’s position, relative to say the origin, is a function, but the derivative of this function is its instantaneous velocity (how fast it’s moving) at a certain point, and the derivative of its velocity is its acceleration (how fast its velocity is changing). Note that the acceleration function is a Higher Order Derivative, since we need to take the derivative of the position function twice to get the acceleration function. You can also think of the instantaneous velocity as the average velocity as the difference in $ x$’s (for example, change in time) gets closer and closer to $ 0$; thus, we are taking a limit, or a derivative.

On the other hand, the average rate of change or average velocity is $ \displaystyle \frac{{\text{change in position}}}{{\text{change in time}}}=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$ from interval $ a$ to $ b$, and you can think of this as $ \displaystyle \text{rate}=\frac{{\text{distance}}}{{\text{time}}}$. These problems typically involve a time interval, as opposed to an instantaneous point. Note that we don’t need calculus to find this – just some algebra!

Again, typically when we are finding a rate of change, or instantaneous rate of change (“IROC”), we will be taking a derivative at a certain point, and when we are finding an average rate of change (“AROC”), we will be using the $ \displaystyle \frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$ (slope) formula over an interval.

Here is a graph to show this; note how the instantaneous change is the slope of a tangent line at a point:
For these problems, find the average rate of change over the interval. Compare this to the instantaneous rate of change at the endpoints of the interval:

Function and Interval	Instantaneous and Average Rates of Change
$ f\left( t \right)=3t-2,\,\,\,\left[ {1,3} \right]$ $ {f}’\left( t \right)=3\,$	Instantaneous rate of change at $ x=1$ and $ x=3$ (endpoints) is the constant 3 at both points since $ {f}’\left( 1 \right)=3$ and $ {f}’\left( 3 \right)=3$. Average rate of change is $ \displaystyle \frac{{f\left( 3 \right)-f\left( 1 \right)}}{{3-1}}=\frac{{\left[ {3\left( 3 \right)-2} \right]-\left[ {3\left( 1 \right)-2} \right]}}{2}=\frac{{7-1}}{2}=3$. Note that the instantaneous rate of change is the same as the average rate of change, since the function is a line, and the slope of the line is always 3.
$ \displaystyle f\left( x \right)=-\frac{2}{x},\,\,\,\,\left[ {2,4} \right]$ $ \begin{array}{l}f\left( x \right)=-2{{x}^{{-1}}}\\{f}’\left( x \right)=2{{x}^{{-2}}}\,\end{array}$	Instantaneous rate of change at $ x=2$ is $ \displaystyle 2{{\left( 2 \right)}^{{-2}}}=\,\,\,2\left( {\frac{1}{{{{2}^{2}}}}} \right)=2\left( {\frac{1}{4}} \right)=\frac{1}{2}$. Instantaneous rate of change at $ x=4$ is $ \displaystyle 2{{\left( 4 \right)}^{{-2}}}=\,\,\,2\left( {\frac{1}{{{{4}^{2}}}}} \right)=\frac{2}{{16}}=\frac{1}{8}$. Average rate of change is $ \displaystyle \frac{{f\left( 4 \right)-f\left( 2 \right)}}{{4-2}}=\frac{{\left( {-\frac{2}{4}} \right)-\left( {-\frac{2}{2}} \right)}}{2}=\frac{{-\frac{1}{2}+1}}{2}=\frac{{\frac{1}{2}}}{2}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$.
$ \displaystyle f\left( \theta \right)=-\cos \theta ,\,\,\,\left[ {0,\frac{\pi }{3}} \right]$ $ {f}’\left( \theta \right)=\sin \theta $	Instantaneous rate of change at $ x=0$ is $ \sin \left( 0 \right)=0$. Instantaneous rate of change at $ \displaystyle x=\frac{\pi }{3}$ is $ \displaystyle \sin \left( {\frac{\pi }{3}} \right)=\frac{{\sqrt{3}}}{2}$. Average rate of change is $ \displaystyle \frac{{f\left( {\frac{\pi }{3}} \right)-f\left( 0 \right)}}{{\frac{\pi }{3}-0}}=\frac{{\left( {-\cos \frac{\pi }{3}} \right)-\,\left( {-\cos 0} \right)}}{{\frac{\pi }{3}}}=\frac{{\left( {-\frac{1}{2}} \right)-\,\left( {-1} \right)}}{{\frac{\pi }{3}}}=\frac{{\frac{1}{2}}}{{\frac{\pi }{3}}}=\frac{1}{2}\cdot \frac{3}{\pi }=\frac{3}{{2\pi }}$.

Function and Interval

Instantaneous and Average Rates of Change

$ f\left( t \right)=3t-2,\,\,\,\left[ {1,3} \right]$

$ {f}’\left( t \right)=3\,$

Instantaneous rate of change at $ x=1$ and $ x=3$ (endpoints) is the constant 3 at both points since $ {f}’\left( 1 \right)=3$ and $ {f}’\left( 3 \right)=3$.

Average rate of change is $ \displaystyle \frac{{f\left( 3 \right)-f\left( 1 \right)}}{{3-1}}=\frac{{\left[ {3\left( 3 \right)-2} \right]-\left[ {3\left( 1 \right)-2} \right]}}{2}=\frac{{7-1}}{2}=3$. Note that the instantaneous rate of change is the same as the average rate of change, since the function is a line, and the slope of the line is always 3.

$ \displaystyle f\left( x \right)=-\frac{2}{x},\,\,\,\,\left[ {2,4} \right]$

$ \begin{array}{l}f\left( x \right)=-2{{x}^{{-1}}}\\{f}’\left( x \right)=2{{x}^{{-2}}}\,\end{array}$

Instantaneous rate of change at $ x=2$ is $ \displaystyle 2{{\left( 2 \right)}^{{-2}}}=\,\,\,2\left( {\frac{1}{{{{2}^{2}}}}} \right)=2\left( {\frac{1}{4}} \right)=\frac{1}{2}$. Instantaneous rate of change at $ x=4$ is $ \displaystyle 2{{\left( 4 \right)}^{{-2}}}=\,\,\,2\left( {\frac{1}{{{{4}^{2}}}}} \right)=\frac{2}{{16}}=\frac{1}{8}$.

Average rate of change is $ \displaystyle \frac{{f\left( 4 \right)-f\left( 2 \right)}}{{4-2}}=\frac{{\left( {-\frac{2}{4}} \right)-\left( {-\frac{2}{2}} \right)}}{2}=\frac{{-\frac{1}{2}+1}}{2}=\frac{{\frac{1}{2}}}{2}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$.

$ \displaystyle f\left( \theta \right)=-\cos \theta ,\,\,\,\left[ {0,\frac{\pi }{3}} \right]$

$ {f}’\left( \theta \right)=\sin \theta $

Instantaneous rate of change at $ x=0$ is $ \sin \left( 0 \right)=0$. Instantaneous rate of change at $ \displaystyle x=\frac{\pi }{3}$ is $ \displaystyle \sin \left( {\frac{\pi }{3}} \right)=\frac{{\sqrt{3}}}{2}$.

Average rate of change is $ \displaystyle \frac{{f\left( {\frac{\pi }{3}} \right)-f\left( 0 \right)}}{{\frac{\pi }{3}-0}}=\frac{{\left( {-\cos \frac{\pi }{3}} \right)-\,\left( {-\cos 0} \right)}}{{\frac{\pi }{3}}}=\frac{{\left( {-\frac{1}{2}} \right)-\,\left( {-1} \right)}}{{\frac{\pi }{3}}}=\frac{{\frac{1}{2}}}{{\frac{\pi }{3}}}=\frac{1}{2}\cdot \frac{3}{\pi }=\frac{3}{{2\pi }}$.

When we’re talking about an object traveling with respect to a time ($ t$) with a position $ s\left( t \right)$, velocity $ v\left( t \right)$, and acceleration $ a\left( t \right)$, here are some concepts. If you’ve taken Physics, you’ve probably dealt with this stuff:

Initially is when $ t=0$, at the origin is when $ s(t)=0$, and at rest is when $ v(t)=0$.
If a particle is moving along a horizontal line, if its position $ s(t)$ (relative to say the origin) is represented by a function, the derivative of this function is its instantaneous velocity $ v(t)$ (how fast it’s moving), and the derivative of its velocity is its acceleration $ a(t)$ (how fast its velocity is changing).
The position of an object is a vector, since it has both a magnitude (a scalar, such as distance) and a direction. A change in position is the displacement vector, which is how far out of place the object is, compared to where it started. The distance it has traveled is the total amount of ground an object has covered during its motion, a scalar, and is the absolute value of the displacement vector.
The velocity function is the derivative of the position function, and can be negative, zero, or positive. If the derivative (velocity) is positive, the object is moving to the right (or up, if that’s how the coordinate system is defined); if negative, it’s moving to the left (or down); if the velocity is 0, the object is at rest. Thus, if the velocity changes sign, the object changes direction.
The velocity of an object is actually a vector, whereas the speed is the absolute value of the velocity vector, and is a scalar. The speed of an object cannot be negative, whereas velocity can.
Total distance traveled is the sum of the absolute values of the differences in positions of all the resting points.
Average Velocity is total displacement during that interval over total time elapsed.
Acceleration (the derivative of velocity, which is also a vector) can cause speed to increase, decrease, or stay the same.
More information about speed (remember that speed is the absolute value of velocity):
- If velocity and acceleration have the same sign, speed is increasing. (Example: a velocity from 40 mph to 45 mph, both positive, or from –40 mph to –45 mph, both negative).
- If velocity and acceleration have opposite signs, speed is decreasing. (Example: a velocity from 45 mph to 40 mph, velocity positive, acceleration negative, or from –45 mph to –40 mph, velocity negative, acceleration positive).
Farthest left (or down) means minimum and farthest right (or up) means maximum.
To find the maximum or minimum velocity or acceleration, find where the derivative of these functions fails or is zero.
The position of free-falling objects is $ \displaystyle s\left( t \right)=\frac{1}{2}g{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$, where $ {{v}_{0}}$ is the initial velocity, $ {{s}_{0}}$ is the initial height, and $ g$ is the acceleration from gravity. On earth, the acceleration due to gravity is about –32 feet per second per second or –9.8 meters per second per second. In algebra, we saw this equation as $ g\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$, since we typically used feet per second. When the object hits the ground, $ s\left( t \right)=0$ (remember to always count the position from the ground up).

Here are some problems using the position, velocity, and acceleration functions:

Position, Velocity and Acceleration Calculus Problem	Solution
A quarter is dropped from a ladder that is 50 feet tall. Use the position function $ s\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ for free falling objects to answer the following ($ t$ is in seconds, $ s$ is in feet): a) What are the position and velocity functions for the coin? b) What is the average velocity of the coin in the $ t$ interval $ [2,3]$? c) What is the instantaneous velocity of the coin at $ t=3$ seconds? d) Find the time when the coin hits the ground, and the velocity at impact.	a) Since there’s no initial velocity (since the coin is dropped from 50 feet), $ {{v}_{0}}=0$ and $ {{h}_{0}}=50$; the position function is $ s\left( t \right)=-16{{t}^{2}}+50$. The velocity function, which is the derivative of the position function, is $ v\left( t \right)=-32t$. b) The average velocity of the quarter in the interval $ [2,3]$ is $ \displaystyle \frac{{f\left( 3 \right)-f\left( 2 \right)}}{{3-2}}=\frac{{\left[ {-16{{{\left( 3 \right)}}^{2}}+50} \right]-\left[ {-16{{{\left( 2 \right)}}^{2}}+50} \right]}}{1}=-80$ feet per second. c) To get the instantaneous velocity of the quarter at $ t=3$, we plug 3 into the velocity function, so we have $ v\left( t \right)=-32t$, or $ v\left( 3 \right)=-32\left( 3 \right)=-96$ feet per second. d) The quarter hits the ground when the height is 0; so $ s\left( t \right)=0$. Thus, $ s\left( t \right)=-16{{t}^{2}}+50=0$, or $ -16{{t}^{2}}=-50$, or $ t\approx 1.77$ seconds. The velocity (instantaneous) at this time is $ v\left( {1.77} \right)=-32\left( {1.77} \right)=-56.6$ feet per second.
A ball is thrown down from the top of a 120-foot bridge with an initial velocity of –20 feet per second. Use the position function $ s\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ for free falling objects to answer the following ($ t$ is in seconds, $ s$ is in feet): a) What is its velocity at after 2 seconds? b) What is its velocity after falling 50 feet?	The position function is $ s\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$, or $ s\left( t \right)=-16{{t}^{2}}-20t+120$. a) To find the (instantaneous) velocity at $ t=2$, take the derivative and plug in 2 for $ t$: $ v\left( t \right)=-32t-20$, so $ v\left( 2 \right)=-32\left( 2 \right)-20=-84$ feet per second. b) To find its velocity after falling 50 feet, first find the time the ball reaches 70 feet from the ground $ (120–50=70)$: $ 70=-16{{t}^{2}}-20t+120$, or $ 0=-16{{t}^{2}}-20t+50$. To solve for $ t$, we can use the quadratic formula or the graphing calculator; we get $ t=1.25$ when the position function is at 70. Now plug 1.25 into the velocity vector to get the (instantaneous) velocity: $ v\left( t \right)=-32t-20$, so $ v\left( {1.25} \right)=-32\left( {1.25} \right)-20=-60$ feet per second.
The velocity function in meters per second is $ v\left( t \right)=40-{{t}^{2}}$ for $ 0\le t\le 5$. a) Find the velocity of the object when $ t=2$. b) Find the acceleration of the object when $ t=2$. c) When the velocity and acceleration have opposite signs, what can be said about the speed of the object?	The velocity function is given at $ v\left( t \right)=40-{{t}^{2}}$. a) To find the velocity at $ t=2$, just plug 2 for $ t$ into the velocity function: $ v\left( 2 \right)=40-{{2}^{2}}=36$ meters per second. b) To find the acceleration at $ t=2$, take the derivative of the velocity function and plug in 2 for $ t$: $ a\left( t \right)={v}’\left( t \right)=-2t$; $ {v}’\left( 2 \right)=-2\cdot 2=-4$ meters per second per second ($ \text{m/se}{{\text{c}}^{2}}$). c) When the velocity and acceleration have opposite signs, the speed is decreasing. This is because if the velocity is positive and the acceleration is negative (like in this case), the speed is getting smaller since it’s slowing down. If the velocity is negative and the acceleration is positive (like from –5 m/sec to –2 m/sec, for example), the speed (absolute value of the velocity) is also decreasing.

Position, Velocity and Acceleration Calculus Problem

Solution

A quarter is dropped from a ladder that is 50 feet tall.

Use the position function $ s\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ for free falling objects to answer the following ($ t$ is in seconds, $ s$ is in feet):

a) What are the position and velocity functions for the coin?

b) What is the average velocity of the coin in the $ t$ interval $ [2,3]$?

c) What is the instantaneous velocity of the coin at $ t=3$ seconds?

d) Find the time when the coin hits the ground, and the velocity at impact.

a) Since there’s no initial velocity (since the coin is dropped from 50 feet), $ {{v}_{0}}=0$ and $ {{h}_{0}}=50$; the position function is $ s\left( t \right)=-16{{t}^{2}}+50$. The velocity function, which is the derivative of the position function, is $ v\left( t \right)=-32t$.

b) The average velocity of the quarter in the interval $ [2,3]$ is $ \displaystyle \frac{{f\left( 3 \right)-f\left( 2 \right)}}{{3-2}}=\frac{{\left[ {-16{{{\left( 3 \right)}}^{2}}+50} \right]-\left[ {-16{{{\left( 2 \right)}}^{2}}+50} \right]}}{1}=-80$ feet per second.

c) To get the instantaneous velocity of the quarter at $ t=3$, we plug 3 into the velocity function, so we have $ v\left( t \right)=-32t$, or $ v\left( 3 \right)=-32\left( 3 \right)=-96$ feet per second.

d) The quarter hits the ground when the height is 0; so $ s\left( t \right)=0$. Thus, $ s\left( t \right)=-16{{t}^{2}}+50=0$, or $ -16{{t}^{2}}=-50$, or $ t\approx 1.77$ seconds. The velocity (instantaneous) at this time is $ v\left( {1.77} \right)=-32\left( {1.77} \right)=-56.6$ feet per second.

A ball is thrown down from the top of a 120-foot bridge with an initial velocity of –20 feet per second.

Use the position function $ s\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ for free falling objects to answer the following ($ t$ is in seconds, $ s$ is in feet):

a) What is its velocity at after 2 seconds?

b) What is its velocity after falling 50 feet?

The position function is $ s\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$, or $ s\left( t \right)=-16{{t}^{2}}-20t+120$.

a) To find the (instantaneous) velocity at $ t=2$, take the derivative and plug in 2 for $ t$: $ v\left( t \right)=-32t-20$, so $ v\left( 2 \right)=-32\left( 2 \right)-20=-84$ feet per second.

b) To find its velocity after falling 50 feet, first find the time the ball reaches 70 feet from the ground $ (120–50=70)$: $ 70=-16{{t}^{2}}-20t+120$, or $ 0=-16{{t}^{2}}-20t+50$. To solve for $ t$, we can use the quadratic formula or the graphing calculator; we get $ t=1.25$ when the position function is at 70. Now plug 1.25 into the velocity vector to get the (instantaneous) velocity: $ v\left( t \right)=-32t-20$, so $ v\left( {1.25} \right)=-32\left( {1.25} \right)-20=-60$ feet per second.

The velocity function in meters per second is $ v\left( t \right)=40-{{t}^{2}}$ for $ 0\le t\le 5$.

a) Find the velocity of the object when $ t=2$.

b) Find the acceleration of the object when $ t=2$.

c) When the velocity and acceleration have opposite signs, what can be said about the speed of the object?

The velocity function is given at $ v\left( t \right)=40-{{t}^{2}}$.

a) To find the velocity at $ t=2$, just plug 2 for $ t$ into the velocity function: $ v\left( 2 \right)=40-{{2}^{2}}=36$ meters per second.

b) To find the acceleration at $ t=2$, take the derivative of the velocity function and plug in 2 for $ t$: $ a\left( t \right)={v}’\left( t \right)=-2t$; $ {v}’\left( 2 \right)=-2\cdot 2=-4$ meters per second per second ($ \text{m/se}{{\text{c}}^{2}}$).

c) When the velocity and acceleration have opposite signs, the speed is decreasing. This is because if the velocity is positive and the acceleration is negative (like in this case), the speed is getting smaller since it’s slowing down. If the velocity is negative and the acceleration is positive (like from –5 m/sec to –2 m/sec, for example), the speed (absolute value of the velocity) is also decreasing.

Here’s one more problem, where we have to distinguish between the instantaneous velocity and the average velocity:

Position, Velocity and Acceleration Calculus Problem	Solution
Suppose a particle is traveling along the $ x$-axis, and its position at time $ t$ is $ \displaystyle s\left( t \right)=\frac{{5{{t}^{2}}}}{{{{t}^{2}}+4}}$, where $ t$ is in seconds. For what value of $ t$ in the interval $ \left[ {2,6} \right]$ is the instantaneous velocity the same as the average velocity?	Since we are given the position, and we need the (instantaneous) velocity, we have to take the derivative (differentiate): $ \displaystyle \begin{align}v\left( t \right)&=\frac{{d\left( {s\left( t \right)} \right)}}{{dt}}=\frac{{d\left( {\frac{{5{{t}^{2}}}}{{{{t}^{2}}+4}}} \right)}}{{dt}}=\frac{{\left( {{{t}^{2}}+4} \right)\,\left( {10t} \right)-\left( {5{{t}^{2}}} \right)\left( {2t} \right)}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}\\&=\frac{{10{{t}^{3}}+40t-10{{t}^{3}}}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}=\frac{{40t}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}\end{align}$ This is the instantaneous velocity at time $ t$. The average velocity is the average rate of change, which is $ \displaystyle \frac{{\text{change in position}}}{{\text{change in time}}}=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$, and we need this for the interval $ [2,6]$. Now, plug in the endpoints: $ \displaystyle \frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}=\frac{{f\left( 6 \right)-f\left( 2 \right)}}{{6-2}}=\frac{{\frac{{5{{{\left( 6 \right)}}^{2}}}}{{{{{\left( 6 \right)}}^{2}}+4}}-\frac{{5{{{\left( 2 \right)}}^{2}}}}{{{{{\left( 2 \right)}}^{2}}+4}}}}{{6-2}}=\frac{{4.5-2.5}}{4}=\frac{1}{2}$ To find out when the instantaneous velocity is the same as the average velocity, we set the two together and solve for $ t$: $ \displaystyle \frac{{40t}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}=\frac{1}{2}$ Cross-multiply to get $ 80t={{t}^{4}}+8{{t}^{2}}+16$ This looks difficult to solve, so let’s use our graphing calculator, with $ \displaystyle {{Y}_{1}}=\frac{{40t}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}$ and $ \displaystyle {{Y}_{2}}=\frac{1}{2}$ (take the intersection) to get $ t\approx 3.602$. Since this is the only intersection in the interval $ [2,6]$, the instantaneous velocity is the same as the average velocity at approximately 3.602 seconds.

Position, Velocity and Acceleration Calculus Problem

Solution

Suppose a particle is traveling along the $ x$-axis, and its position at time $ t$ is $ \displaystyle s\left( t \right)=\frac{{5{{t}^{2}}}}{{{{t}^{2}}+4}}$, where $ t$ is in seconds.

For what value of $ t$ in the interval $ \left[ {2,6} \right]$ is the instantaneous velocity the same as the average velocity?

Since we are given the position, and we need the (instantaneous) velocity, we have to take the derivative (differentiate):

$ \displaystyle \begin{align}v\left( t \right)&=\frac{{d\left( {s\left( t \right)} \right)}}{{dt}}=\frac{{d\left( {\frac{{5{{t}^{2}}}}{{{{t}^{2}}+4}}} \right)}}{{dt}}=\frac{{\left( {{{t}^{2}}+4} \right)\,\left( {10t} \right)-\left( {5{{t}^{2}}} \right)\left( {2t} \right)}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}\\&=\frac{{10{{t}^{3}}+40t-10{{t}^{3}}}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}=\frac{{40t}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}\end{align}$

This is the instantaneous velocity at time $ t$.

The average velocity is the average rate of change, which is $ \displaystyle \frac{{\text{change in position}}}{{\text{change in time}}}=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$, and we need this for the interval $ [2,6]$. Now, plug in the endpoints:

$ \displaystyle \frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}=\frac{{f\left( 6 \right)-f\left( 2 \right)}}{{6-2}}=\frac{{\frac{{5{{{\left( 6 \right)}}^{2}}}}{{{{{\left( 6 \right)}}^{2}}+4}}-\frac{{5{{{\left( 2 \right)}}^{2}}}}{{{{{\left( 2 \right)}}^{2}}+4}}}}{{6-2}}=\frac{{4.5-2.5}}{4}=\frac{1}{2}$

To find out when the instantaneous velocity is the same as the average velocity, we set the two together and solve for $ t$:

$ \displaystyle \frac{{40t}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}=\frac{1}{2}$ Cross-multiply to get $ 80t={{t}^{4}}+8{{t}^{2}}+16$

This looks difficult to solve, so let’s use our graphing calculator, with $ \displaystyle {{Y}_{1}}=\frac{{40t}}{{{{{\left( {{{t}^{2}}+4} \right)}}^{2}}}}$ and $ \displaystyle {{Y}_{2}}=\frac{1}{2}$ (take the intersection) to get $ t\approx 3.602$. Since this is the only intersection in the interval $ [2,6]$, the instantaneous velocity is the same as the average velocity at approximately 3.602 seconds.

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