Implicit Differentiation and Related Rates

Introduction to Implicit Differentiation	Using Implicit Differentiation to Find Higher Order Derivatives
Equation of the Tangent Line with Implicit Differentiation	Related Rates
More Practice

Introduction to Implicit Differentiation

Up to now, we’ve differentiated in explicit form, since, for example, $ y$ has been explicitly written as a function of $ x$. Sometimes we can’t get an equation with a “$ y$” only on one side; we may have multiply “$ y$’s” in the equation. In these cases, we have to differentiate “implicitly”, meaning that some “$ y$’s” are “inside” the equation. This is called implicit differentiation, and we actually have to use the the Chain Rule to do this. Here’s an example of an equation that we’d have to differentiate implicitly: $ y=7{{x}^{2}}y-2{{y}^{2}}-\sqrt{{xy}}$. Do you see how it’d be really difficult to get $ y$ alone on one side?

The main thing to remember is when you are differentiating with respect to “$ x$” and what you are differentiating only has “$ \boldsymbol {x}$’s” in it (or constants), you just get the derivative the normal way (since the $ dx$’s cancel out). But if you are differentiating with respect to $ x$, and what you are differentiating has another variable in it, like “$ y$”, you have to multiply by $ \displaystyle \frac{{dy}}{{dx}}\text{ (}{y}’)$:

$ \require{cancel} \displaystyle \frac{d}{{dx}}\left( {{{x}^{2}}} \right)=2x\cdot \frac{{\cancel{{dx}}}}{{\cancel{{dx}}}}=2x$: variables agree, so just use the power rule (chain rule has no effect)

$ \displaystyle \frac{d}{{dx}}\left( {{{y}^{2}}} \right)=2y\cdot \frac{{dy}}{{dx}}=2y{y}’$: variables disagree, so use the power rule, and then the chain rule

After we do the differentiation, we want to solve for the $ \displaystyle \frac{{dy}}{{dx}}$ (also written as $ {y}’$) by getting it to one side by itself (and we may have both $ x$’s and $ y$’s on the other side, which is fine). Here are the steps for differentiating implicitly:

Implicit Differentiation

Differentiate both sides of equation with respect to $ \boldsymbol{x}$. When are you differentiating variables other than $ x$ (such as “$ y$”), remember to multiply that term by $ \displaystyle \frac{{dy}}{{dx}}\text{ (}{y}’)$!
Move $ {y}’$ to the left side of the equation and move all other terms to the right side (even if they have $ x$’s and $ y$’s in them).
Factor out the $ {y}’$ on the left side of the equation, and solve for $ {y}’$.

Example: $ \displaystyle \begin{align}{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=\frac{{-2x}}{{2y}}=-\frac{x}{y}\end{align}$

I know it looks a bit scary, but it’s really not that bad! Let’s do some non-trig problems first. Do you see how we have to use the chain rule a lot more? And note how the algebra can get really complicated (and also, the answer can be in different forms)!

Implicit Differentiation Problem	Solution
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \left( {x-3} \right)\left( {y+5} \right)=17$	Using the power rule first: $ \displaystyle \begin{align}\left( {x-3} \right)\left( {y+5} \right)&=17\\\left( {x-3} \right){{\left( {y+5} \right)}^{\prime }}+\left( {y+5} \right){{\left( {x-3} \right)}^{\prime }}&=0\\\left( {x-3} \right)\cdot 1{y}’+\left( {y+5} \right)\cdot 1&=0\\{y}’&=\frac{{-\left( {y+5} \right)}}{{x-3}}=\frac{{y+5}}{{3-x}}\end{align}$ Multiplying binomials first: $ \displaystyle \begin{align}\left( {x-3} \right)\left( {y+5} \right)&=17\\xy+5x-3y-15&=17\\\left( {x{y}’+y\cdot 1} \right)+5\cdot 1-3{y}’-0&=0\\{y}’\left( {x-3} \right)&=-y-5\\{y}’&=\frac{{-y-5}}{{x-3}}=\frac{{y+5}}{{3-x}}\end{align}$
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \displaystyle \frac{{x+y}}{{{{y}^{2}}}}=xy$	$ \displaystyle \begin{align}\frac{{x+y}}{{{{y}^{2}}}}&=xy\\\frac{{{{y}^{2}}\left( {1+{y}’} \right)-\left( {x+y} \right)\left( {2y{y}’} \right)}}{{{{y}^{4}}}}&=x{y}’+y\cdot 1\,\,\,\,\,\,\text{quotient rule/product rule}\\{{y}^{2}}+{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’&=\left( {x{y}’+y} \right){{y}^{4}}\\{{y}^{2}}+{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’&=x{{y}^{4}}{y}’+{{y}^{5}}\\{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’-x{{y}^{4}}{y}’&={{y}^{5}}-{{y}^{2}}\\{y}’\left( {{{y}^{2}}-2xy-2{{y}^{2}}-x{{y}^{4}}} \right)&={{y}^{5}}-{{y}^{2}}\\{y}’&=\frac{{{{y}^{5}}-{{y}^{2}}}}{{-{{y}^{2}}-2xy-x{{y}^{4}}}}=\frac{{{{y}^{4}}-y}}{{-y-2x-x{{y}^{3}}}}\end{align}$
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \sqrt{{xy}}+2x{{y}^{2}}=3$	$ \displaystyle \begin{align}{{\left( {xy} \right)}^{{\frac{1}{2}}}}+2x{{y}^{2}}&=3\\\frac{1}{2}{{\left( {xy} \right)}^{{-\frac{1}{2}}}}\left( {x{y}’+y\cdot 1} \right)+2\left( {x\cdot 2y{y}’+{{y}^{2}}\cdot 1} \right)&=0\\\cancel{{{{{\left( {xy} \right)}}^{{\frac{1}{2}}}}}}\left[ {\frac{1}{2}\cancel{{{{{\left( {xy} \right)}}^{{-\frac{1}{2}}}}}}\left( {x{y}’+y} \right)+2\left( {2xy{y}’+{{y}^{2}}} \right)} \right]&=0\cdot {{\left( {xy} \right)}^{{\frac{1}{2}}}}\,\,\,\,\text{multiply both sides by }{{\left( {xy} \right)}^{{\frac{1}{2}}}}\\\frac{1}{2}\left( {x{y}’+y} \right)+4xy{y}'{{\left( {xy} \right)}^{{\frac{1}{2}}}}+2{{y}^{2}}{{\left( {xy} \right)}^{{\frac{1}{2}}}}&=0\\x{y}’+y+8xy{y}'{{\left( {xy} \right)}^{{\frac{1}{2}}}}+4{{y}^{2}}{{\left( {xy} \right)}^{{\frac{1}{2}}}}&=0\,\,\,\,\,\text{multiply both sides by }2\\{y}’\left( {8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x} \right)&=-y-4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}\\{y}’&=\frac{{-y-4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}}}{{8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x}}=-\frac{{y+4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}}}{{8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x}}\end{align}$

Let’s try some problems with trig now. Note that with trig functions, it’s hard to know where to stop simplifying, so there are several “correct” answers.

Implicit Differentiation Problem	Solution
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \sin x+4\cos \left( {3y} \right)=1$	$ \displaystyle \begin{align}\sin x+4\cos \left( {3y} \right)&=1\\\cos x+4\cdot -\sin \left( {3y} \right)\cdot 3\cdot {y}’&=0\\-12{y}’\sin \left( {3y} \right)&=-\cos x\\{y}’&=\frac{{-\cos x}}{{-12\sin \left( {3y} \right)}}=\frac{{\cos x}}{{12\sin \left( {3y} \right)}}=\frac{1}{{12}}\cos x\csc \left( {3y} \right)\end{align}$
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \cos x=x\left( {2+\csc y} \right)$	$ \displaystyle \begin{align}\cos x&=x\left( {2+\csc y} \right)\\-\sin x&=x\left( {0+-\csc y\cot y\cdot {y}’} \right)+\left( {2+\csc y} \right)\left( 1 \right)\\-\sin x&=-x{y}’\csc y\cot y+2+\csc y\\x{y}’\csc y\cot y&=2+\csc y+\sin x\\{y}’&=\frac{{2+\csc y+\sin x}}{{x\csc y\cot y}}\\&=\frac{2}{{x\csc y\cot y}}+\frac{{\cancel{{\csc y}}}}{{x\cancel{{\csc y}}\cot y}}+\frac{{\sin x}}{{x\csc y\cot y}}\,\,\,\,\,\,(\text{expand)}\\&=\frac{2}{{x\frac{1}{{\sin y}}\cdot \frac{{\cos y}}{{\sin y}}}}+\frac{1}{{x\cdot \frac{{\cos y}}{{\sin y}}}}+\frac{{\sin x}}{{x\cdot \frac{1}{{\sin y}}\cdot \frac{{\cos y}}{{\sin y}}}}\,\,\,\,\,\,\,\text{(put in terms of sin,cos)}\\&=\frac{{2{{{\sin }}^{2}}y}}{{x\cos y}}+\frac{{\tan y}}{x}+\frac{{\sin x{{{\sin }}^{2}}y}}{{x\cos y}}=\frac{{2\sin y\sin y}}{{x\cos y}}+\frac{{\tan y}}{x}+\frac{{\sin x\sin y\sin y}}{{x\cos y}}\\&=\frac{{2\sin y\tan y+\tan y+\sin x\sin y\tan y}}{x}\end{align}$ (see how we can get carried away with simplifying?)
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \displaystyle \begin{array}{c}{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{2}}=4\end{array}$	$ \displaystyle \begin{align}{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{2}}&=4\\2{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{1}}\left( {\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{{\sec }}^{2}}\left( {\pi y} \right)\cdot \pi {y}’} \right)&=0\\\frac{{\cancel{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}\left( {\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{{\sec }}^{2}}\left( {\pi y} \right)\cdot \pi {y}’} \right)}}{{\cancel{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}}}&=\frac{0}{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}\\\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{\sec }^{2}}\left( {\pi y} \right)\cdot \pi {y}’&=0\\{{\sec }^{2}}\left( {\pi y} \right)\cdot \pi {y}’&=-\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi \\{y}’&=\frac{{-\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \cancel{\pi }}}{{{{{\sec }}^{2}}\left( {\pi y} \right) \cancel{\pi }}}\\=-\sec \left( {\pi x} \right)\tan &\left( {\pi x} \right){{\cos }^{2}}\left( {\pi y} \right)\end{align}$
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation: $ \displaystyle \cos (xy)=5x$	$ \displaystyle \begin{align}\cos \left( {xy} \right)&=5x\\-\sin \left( {xy} \right)\left( {x{y}’+y\cdot 1} \right)&=5\\\sin \left( {xy} \right)\left( {x{y}’+y} \right)&=-5\\x{y}’\sin \left( {xy} \right)+y\sin \left( {xy} \right)=-5\\x{y}’\sin \left( {xy} \right)&=-5-y\sin \left( {xy} \right)\\{y}’&=-\frac{{5+y\sin \left( {xy} \right)}}{{x\sin \left( {xy} \right)}}=\frac{{-5\csc \left( {xy} \right)+y}}{x}\end{align}$

Using Implicit Differentiation to Find Higher Order Derivatives

Here’s a problem where we have to use implicit differentiation twice to find the second derivative $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$, or $ {y}^{\prime \prime}$. This one is really tricky, since we need to substitute what we got for $ \displaystyle \frac{{dy}}{{dx}}\,\,or\,\,{y}’$ to simplify.

Note that in this example, we also substituted the original function back in to simplify further. (The reason I substituted at the end is because the resulting answer was one of the choices on a multiple-choice test question. Your answer could be in a number of different forms.)

Implicit Differentiation Problem	Solution
Use implicit differentiation to find $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$: $ {{y}^{5}}={{x}^{{10}}}$ (Note that this is the same as $ \displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1$)	$ \displaystyle \begin{align}{{y}^{5}}&={{x}^{{10}}}\\5{{y}^{4}}{y}’&=10{{x}^{9}}\\{y}’&=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\end{align}$ $ \displaystyle {y}’=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\times \frac{{xy}}{{xy}}\,=\frac{{10{{x}^{{10}}}y}}{{5x{{y}^{5}}}}=\frac{{10y}}{{5x}}\times \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=\frac{{10y}}{{5x}}\times 1\,=\frac{{2y}}{x}$ Notice the cool trick to simplify before taking the second derivative. Since we ended up with $ \displaystyle \frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}$ after differentiating once, and we know from the original equation that $ \displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1$, we can multiply by $ \displaystyle \frac{{xy}}{{xy}}$ to substitute and simplify!
Now take the second derivative using implicit differentiation and the quotient rule: $ \require{cancel} \displaystyle \begin{align}{y}’&=\frac{{2y}}{x}\\{y}^{\prime \prime}&=\frac{{x\cdot 2{y}’-2y\left( 1 \right)}}{{{{x}^{2}}}}\\&=\frac{{\cancel{x}\cdot 2\left( {\frac{{2y}}{{\cancel{x}}}} \right)-2y}}{{{{x}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,(\text{plug}\,\text{ }{y}’\,\,\text{back in)}\\&=\frac{{4y-2y}}{{{{x}^{2}}}}=\frac{{2y}}{{{{x}^{2}}}}\end{align}$

Implicit Differentiation Problem

Solution

Use implicit differentiation to find $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$:

$ {{y}^{5}}={{x}^{{10}}}$

(Note that this is the same as $ \displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1$)

$ \displaystyle \begin{align}{{y}^{5}}&={{x}^{{10}}}\\5{{y}^{4}}{y}’&=10{{x}^{9}}\\{y}’&=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\end{align}$ $ \displaystyle {y}’=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\times \frac{{xy}}{{xy}}\,=\frac{{10{{x}^{{10}}}y}}{{5x{{y}^{5}}}}=\frac{{10y}}{{5x}}\times \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=\frac{{10y}}{{5x}}\times 1\,=\frac{{2y}}{x}$

Notice the cool trick to simplify before taking the second derivative. Since we ended up with $ \displaystyle \frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}$ after differentiating once, and we know from the original equation that $ \displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1$, we can multiply by $ \displaystyle \frac{{xy}}{{xy}}$ to substitute and simplify!

Now take the second derivative using implicit differentiation and the quotient rule:

$ \require{cancel} \displaystyle \begin{align}{y}’&=\frac{{2y}}{x}\\{y}^{\prime \prime}&=\frac{{x\cdot 2{y}’-2y\left( 1 \right)}}{{{{x}^{2}}}}\\&=\frac{{\cancel{x}\cdot 2\left( {\frac{{2y}}{{\cancel{x}}}} \right)-2y}}{{{{x}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,(\text{plug}\,\text{ }{y}’\,\,\text{back in)}\\&=\frac{{4y-2y}}{{{{x}^{2}}}}=\frac{{2y}}{{{{x}^{2}}}}\end{align}$

Equation of the Tangent Line with Implicit Differentiation

Here are some problems where you have to use implicit differentiation to find the derivative at a certain point, and the slope of the tangent line to the graph at a certain point. The last problem asks to find the equation of the tangent line and normal line (the line perpendicular to the tangent line; thus, taking the negative reciprocal of the slope) at a certain point.

Note that we learned about finding the Equation of the Tangent Line in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change section.

Implicit Differentiation Tangent Problem	Solution
Find $ \displaystyle \frac{{dx}}{{dy}}$ by implicit differentiation and evaluate the derivative at point $ \left( {-2,4} \right)$: $ {{x}^{2}}+{{y}^{3}}=68$	$ \displaystyle \begin{align}{{x}^{2}}+{{y}^{3}}&=68\\2x+3{{y}^{2}}{y}’&=0\\3{{y}^{2}}{y}’&=-2x\\{y}’&=-\frac{{2x}}{{3{{y}^{2}}}}\end{align}$ $ \displaystyle \text{At }\left( {-2,4} \right),\,{y}’=-\frac{{2\left( {-2} \right)}}{{3{{{\left( 4 \right)}}^{2}}}}=-\frac{{-4}}{{48}}=\frac{1}{{12}}$
Find the slope of the tangent line to the graph at point $ \left( {2,1} \right)$: $ \displaystyle {{x}^{3}}+{{y}^{3}}=\frac{9}{2}xy$	$ \displaystyle \begin{align}{{x}^{3}}+{{y}^{3}}&=\frac{9}{2}xy\\3{{x}^{2}}+3{{y}^{2}}{y}’&=\frac{9}{2}\left( {x{y}’+y\cdot 1} \right)\\6{{x}^{2}}+6{{y}^{2}}{y}’&=9x{y}’+9y\\6{{y}^{2}}{y}’-9x{y}’&=-6{{x}^{2}}+9y\\{y}’\left( {6{{y}^{2}}-9x} \right)&=-6{{x}^{2}}+9y\\{y}’&=\frac{{-6{{x}^{2}}+9y}}{{6{{y}^{2}}-9x}}=\frac{{-\cancel{3}\left( {2{{x}^{2}}-3y} \right)}}{{\cancel{3}\left( {2{{y}^{2}}-3x} \right)}}\\{y}’&=-\frac{{2{{x}^{2}}-3y}}{{2{{y}^{2}}-3x}}=\frac{{2{{x}^{2}}-3y}}{{3x-2{{y}^{2}}}}\end{align}$ $ \displaystyle \text{At }\left( {2,1} \right),\,\,\,{y}’=\frac{{2{{{\left( 2 \right)}}^{2}}-3\left( 1 \right)}}{{3\left( 2 \right)-2{{{\left( 1 \right)}}^{2}}}}=\frac{5}{4}$
Find the equation of for the tangent line and the normal line to the circle at the point $ \left( {3,4} \right)$: $ {{x}^{2}}+{{y}^{2}}=25$	$ \displaystyle \begin{align}\\{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}$ $ \displaystyle \begin{align}\text{Tangent Line:}\\\text{At }\left( {3,4} \right),\,{y}’=-\frac{3}{4}\\y-{{y}_{1}}=-\frac{3}{4}\left( {x-{{x}_{1}}} \right)\\y-4=-\frac{3}{4}\left( {x-3} \right)\\y=-\frac{3}{4}x+\frac{{25}}{4}\end{align}$ $ \displaystyle \begin{align}\text{Normal Line:}\\\text{(slope is negative reciprocal)}\\\text{At }\left( {3,4} \right),\,\,\,{y}’=-\frac{3}{4}\\y-{{y}_{1}}=\frac{4}{3}\left( {x-{{x}_{1}}} \right)\\y-4=\frac{4}{3}\left( {x-3} \right)\\y=\frac{4}{3}x\end{align}$

Here’s a problem where we have to use implicit differentiation to find the points at which the graph of the equation has a horizontal and vertical tangent line. Remember that for the horizontal tangent line, we set the numerator to 0, and for the vertical tangent line, we set the denominator to 0.

Implicit Differentiation and Horizontal/Vertical Tangent Lines
Use implicit differentiation to find the points at which the graph of the equation has a horizontal and vertical tangent line: $ 2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0$
Use Implicit Differentiation to get $ {y}’$: $ \displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\4x+2y{y}’-4+2{y}’+0=0\\2y{y}’+2{y}’=-4x+4\\{y}’\left( {2y+2} \right)=-4x+4\end{array}$ $ \displaystyle {y}’=\frac{{-4x+4}}{{2y+2}}$ $ \displaystyle {y}’=\frac{{2-2x}}{{y+1}}$	Points at Horizontal Tangent (set numerator to 0): $ \displaystyle \begin{array}{c}2-2x=0\\x=1\end{array}$ Now find $ y$ (use original): $ \displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{\left( 1 \right)}^{2}}+{{y}^{2}}-4\left( 1 \right)+2y+2=0\\{{y}^{2}}+2y=0\\y\left( {y+2} \right)=0\\y=0,\,-2\\\left( {1,0} \right)\,\,\text{and}\,\,\left( {1,-2} \right)\end{array}$	Points at Vertical Tangent (set denominator to 0): $ \displaystyle \begin{array}{c}y+1=0\\y=-1\end{array}$ Now find $ x$ (use original): $ \displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{x}^{2}}+{{\left( {-1} \right)}^{2}}-4x+2\left( {-1} \right)+2=0\\2{{x}^{2}}-4x+1=0\end{array}$ $ \displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{-\left( {-4} \right)\pm \sqrt{{{{{\left( {-4} \right)}}^{2}}-4\cdot 2\cdot 1}}}}{{2\cdot 2}}$ $ \displaystyle x=\frac{{4\pm \sqrt{8}}}{4}=\frac{{4\pm 2\sqrt{2}}}{4}=\frac{{2\pm \sqrt{2}}}{2}$ $ \displaystyle \left( {\frac{{2-\sqrt{2}}}{2},-1} \right)\,\,\text{and}\,\,\left( {\frac{{2+\sqrt{2}}}{2},-1} \right)$

Implicit Differentiation and Horizontal/Vertical Tangent Lines

Use implicit differentiation to find the points at which the graph of the equation has a horizontal and vertical tangent line:

$ 2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0$

Use Implicit Differentiation to get $ {y}’$:

$ \displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\4x+2y{y}’-4+2{y}’+0=0\\2y{y}’+2{y}’=-4x+4\\{y}’\left( {2y+2} \right)=-4x+4\end{array}$

$ \displaystyle {y}’=\frac{{-4x+4}}{{2y+2}}$

$ \displaystyle {y}’=\frac{{2-2x}}{{y+1}}$

Points at Horizontal Tangent (set numerator to 0):

$ \displaystyle \begin{array}{c}2-2x=0\\x=1\end{array}$

Now find $ y$ (use original):

$ \displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{\left( 1 \right)}^{2}}+{{y}^{2}}-4\left( 1 \right)+2y+2=0\\{{y}^{2}}+2y=0\\y\left( {y+2} \right)=0\\y=0,\,-2\\\left( {1,0} \right)\,\,\text{and}\,\,\left( {1,-2} \right)\end{array}$

Points at Vertical Tangent (set denominator to 0):

$ \displaystyle \begin{array}{c}y+1=0\\y=-1\end{array}$

Now find $ x$ (use original):

$ \displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{x}^{2}}+{{\left( {-1} \right)}^{2}}-4x+2\left( {-1} \right)+2=0\\2{{x}^{2}}-4x+1=0\end{array}$

$ \displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{-\left( {-4} \right)\pm \sqrt{{{{{\left( {-4} \right)}}^{2}}-4\cdot 2\cdot 1}}}}{{2\cdot 2}}$

$ \displaystyle x=\frac{{4\pm \sqrt{8}}}{4}=\frac{{4\pm 2\sqrt{2}}}{4}=\frac{{2\pm \sqrt{2}}}{2}$

$ \displaystyle \left( {\frac{{2-\sqrt{2}}}{2},-1} \right)\,\,\text{and}\,\,\left( {\frac{{2+\sqrt{2}}}{2},-1} \right)$

Related Rates

I used to have such a problem with related rates problems, until I began writing down the steps to do them. It also helps to remember that the rates in these problems typically are differentiated with respect to time, or $ \displaystyle \frac{{d\left( {\text{something}} \right)}}{{dt}}$. Hope these steps help:

Draw a picture (if applicable) and label any amounts that could be changing. These amounts will be variables, even if the problem gives you an amount at a certain time for these parts (the “when”). In other words, understand what is changing and what isn’t. I find it helpful to draw an arrow to show where the picture is “moving”.
On the picture you’ve drawn, label places that are changing (you’ll need variables), and places that aren’t changing (constants). For constants, you can put numbers on the picture, if they are given.
Write down exactly what the problem gives you, and what you need to solve for. For example, you may write down “Find $ \displaystyle \frac{{d\left( {\text{Area}} \right)}}{{dt}}$, when $ \text{radius}=6$”. Remember that the rates (things that are changing) have “$ dt$” (with respect to time) in them.$
For the rates, remember that rates with words like “filling up” means a positive volume rate, “emptying out” means a negative volume rate.
Now we have to figure out a way to relate all the variables together. Write an equation that relates all of the given information and variables that you’ve written down. A lot of times this is a geometry equation for volume, area, or perimeter. Also, you’ll see the Pythagorean Equation or the trigonometric functions, such as SOH-CAH-TOA. Remember that shapes stay in proportion, so we may have to use geometrically similar figures and set up proportions. Other constants can be solved for any instant of time; for example, if you have “when length is 40“, you can get the width at that moment too if you can, to use as a constant.
Simplify by trying to put everything in as few variables as possible before differentiating (you may have to substitute some variables by solving in terms of other variables). For example, you may need to relate length and width together if you have a perimeter. I find this the hardest part – how to know what to plug in before I differentiate, and what to plug in after. Other constants can be solved for any instant of time; for example, if you have “when length is $ 40$”, you can get the width at that moment too, to use as a constant. Remember that every variable you have before differentiating will eventually have a “$ \boldsymbol {dt}$” denominator after differentiating, so get rid of any variables where rates of that variable aren’t given or needed in the problem!
The rule of thumb is that when you have values that say “when” something happens, these are put in after differentiating.
Use implicit differentiation to differentiate with respect to time. Hopefully the rates ($ \displaystyle \frac{{d\left( {\text{something}} \right)}}{{dt}}$) you have left at this point are rates that either you have values for, or need values for.
Fill in the equation with what you know (for example, the “when”s). Solve the equation for what you need (what you don’t know). Make sure units match!
Make sure you are answering the question that is being asked, and the units are correct. You’d hate to do all this work, and then answer the wrong question. Remember when dealing with angles, make sure the calculator is in the correct mode (radians or degrees)!
Types of problems:
1. Geometric Shapes typically use area and volume
2. Shadow problems typically use similar triangles
3. Right Triangle problems (such as Ladder problems) typically use the Pythagorean Theorem
4. Problems with Angles typically use Trigonometry

Here are some problems:

Related Rates Problem	Steps and Solution
A point moves along the curve $ y=3{{x}^{2}}-4$ in such a way that the $ y$-value is decreasing at a rate of 3 units per second. At what rate is x changing when $ \displaystyle x=\frac{1}{5}$?	We don’t need to draw a picture, since there’s no application Define what we have and what we want when. We have: $ \displaystyle \frac{{dy}}{{dt}}=-3$, since the $ y$-value is decreasing. We want: find $ \displaystyle \frac{{dx}}{{dt}}$ when $ \displaystyle x=\frac{1}{5}$. Determine what equation we need. Us the given equation $ y=3{{x}^{2}}-4$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time. $ \displaystyle y=3{{x}^{2}}-4;\,\,\,\frac{{dy}}{{dt}}=6x\frac{{dx}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need. $ \displaystyle \frac{{dy}}{{dt}}=6x\frac{{dx}}{{dt}};\,\,\,-3=6\left( {\frac{1}{5}} \right)\frac{{dx}}{{dt}};\,\,\,\frac{{dx}}{{dt}}=-2.5$. Make sure we’re answering the question. Yes, $ -2.5$ is the rate $ x$ is changing.
A 40-foot ladder is resting against a wall, and its base is slipping on the floor (away from the wall) at a rate of 2 feet per minute. Find the rate of change of the height of the ladder at the time when the base is 20 feet from the base of the wall.	Draw a picture. Note that since height ($ y$) and distance from base ($ x$) are changing, we have to use variables for them. The ladder (slanted) will be the hypotenuse of a right triangle formed by the wall, the floor, and the ladder. Define what we have and what we want when. We have: $ \displaystyle \frac{{dx}}{{dt}}=2$. We want: find $ \displaystyle \frac{{dy}}{{dt}}$ when $ x=20$. Determine what equation we need. We need to relate the variables together; use the Pythagorean Theorem: $ {{x}^{2}}+{{y}^{2}}={{40}^{2}}$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time: $ \displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0$. Plug in what we know at the time we know it, and solve for what we need. The tricky part here is that we need to use the Pythagorean Theorem again to get what $ y$ (height) is when $ x=20$ (distance from base of wall): $ {{20}^{2}}+{{y}^{2}}={{40}^{2}};\,\,\,y=\sqrt{{1200}}\approx 34.641$. Now plug everything in: $ \displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0;\,\,\,\,2\left( {20} \right)\left( 2 \right)+2\left( {34.641} \right)\frac{{dy}}{{dt}}=0$; $ \displaystyle \frac{{dy}}{{dt}}=-\frac{{2\left( {20} \right)\left( 2 \right)}}{{2\left( {34.641} \right)}}\approx -1.155\,\,\text{ft/min}$. Make sure we’re answering the question. Yes, $ -1.155$ feet per minute is the rate the height of the ladder is changing. It makes sense that it is negative, since the ladder is slipping down the wall.
The width of a rectangle is increasing at a rate of 10 inches per second, and the area is increasing at a rate of 600 inches² per second. When the width is 25 inches and the area is 1000 inches², how fast is the length of the rectangle changing?	Draw a picture. Note that since width ($ w$) and area ($ A$) are changing, we have to use variables for them. The length ($ l$) is also changing. Define what we have and what we want when. We have: $ \displaystyle \frac{{dw}}{{dt}}=10$ and $ \displaystyle \frac{{dA}}{{dt}}=600$. We want: find $ \displaystyle \frac{{dl}}{{dt}}$ when $ w=25$ and $ A=1000$. Determine what equation we need. We need to relate the variables together; use the area of a triangle: $ A=l\times w$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time: Use product rule: $ \displaystyle A=l\times w;\,\,\,\frac{{dA}}{{dt}}=l\cdot \frac{{dw}}{{dt}}+w\cdot \frac{{dl}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need. The tricky part here is that we need to use the Area of a Rectangle again to get what $ l$ (length) is when $ w=25$ and $ A=1000$: $ A=l\times w;\,\,1000=l\times 25;\,\,l=40$. Now plug everything in: $ \displaystyle \frac{{dA}}{{dt}}=l\cdot \frac{{dw}}{{dt}}+w\cdot \frac{{dl}}{{dt}};\,\,\,600=40\cdot 10+25\cdot \frac{{dl}}{{dt}};\,\,\frac{{dl}}{{dt}}=8\,\,\text{in/min}$. Make sure we’re answering the question. Yes, $ 8$ inches per minute is the rate the length is changing. It makes sense that it is positive, since the rectangle is getting larger.
A woman 5 feet tall walks at a rate of 6 feet per second away from a lamppost. When the woman is 6 feet from the lamppost, her shadow is 8 feet tall. At what rate is the length of her shadow changing when she is 12 feet from the lamppost? At what rate is the tip of her shadow changing when she is 12 feet from the lamppost?	Draw a picture. Remember that the woman’s shadow is on the ground in front of her and gets bigger as she moves away from the light; we’ll use $ y$ for this distance. The distance from the base of the lamppost to the woman is $ x$. We need variables for both of these distances, since they are changing. Define what we have and what we want when. The rate the woman walks from the lamppost, $ \displaystyle \frac{{dx}}{{dt}}=6$. Now this is tricky: to get the rate the length of her shadow increases, we want $ \displaystyle \frac{{dy}}{{dt}}$, but if we want the rate the tip of her shadow increases, we want $ \displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}$. (This is because we need to measure the tip of the shadow in reference to the base of the lamppost). We also have the height of the woman (5 feet), and we know that when the woman is 6 feet from the lamppost $ (x=6),$ her shadow is 8 feet ($ y=8$). Since we will be using similar triangles to relate the triangles together before differentiating (no other way!), we will need the height $ h$ of the lamppost. We can get this using similar triangles when she is 6 feet away from the lamppost (see last triangle to the left): $ \displaystyle \frac{{\text{side of large }\Delta }}{{\text{side of small }\Delta }}=\frac{{\text{bottom of large }\Delta }}{{\text{bottom of small }\Delta }}:\,\,\,\,\,\,\frac{h}{5}=\frac{{14}}{8};\,\,h=8.75$. Determine what equation we need. We need to relate $ x$ and $ y$ to what we know ($ h$ and woman’s height), which we can do using similar triangles again: $ \displaystyle \frac{{\text{8}\text{.75}}}{5}=\frac{{x+y}}{y},$ or (simplified), $ 3.75y=5x$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time. $ \displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need. Now plug in: $ \displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}};\,\,\,\,3.75\frac{{dy}}{{dt}}=5\left( 6 \right);\,\,\,\frac{{dy}}{{dt}}=8$ feet per second. This is the rate of the length of the shadow. To get the rate of the tip of the shadow, we need $ \displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}$, which is actually $ \displaystyle \frac{{dx}}{{dt}}+\frac{{dy}}{{dt}}$, by the sum rule (derivative of a sum is equal to the sum of the derivatives). The rate of the tip of the shadow is $ 6+8=14$ feet per second. (Note that could have also defined a variable to represent $ x + y$, and do some subtraction in setting it up).

Related Rates Problem

Steps and Solution

A point moves along the curve $ y=3{{x}^{2}}-4$ in such a way that the $ y$-value is decreasing at a rate of 3 units per second.

At what rate is x changing when $ \displaystyle x=\frac{1}{5}$?

We don’t need to draw a picture, since there’s no application

Define what we have and what we want when. We have: $ \displaystyle \frac{{dy}}{{dt}}=-3$, since the $ y$-value is decreasing. We want: find $ \displaystyle \frac{{dx}}{{dt}}$ when $ \displaystyle x=\frac{1}{5}$.
Determine what equation we need. Us the given equation $ y=3{{x}^{2}}-4$. Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time. $ \displaystyle y=3{{x}^{2}}-4;\,\,\,\frac{{dy}}{{dt}}=6x\frac{{dx}}{{dt}}$.
Plug in what we know at the time we know it, and solve for what we need. $ \displaystyle \frac{{dy}}{{dt}}=6x\frac{{dx}}{{dt}};\,\,\,-3=6\left( {\frac{1}{5}} \right)\frac{{dx}}{{dt}};\,\,\,\frac{{dx}}{{dt}}=-2.5$.
Make sure we’re answering the question. Yes, $ -2.5$ is the rate $ x$ is changing.

A 40-foot ladder is resting against a wall, and its base is slipping on the floor (away from the wall) at a rate of 2 feet per minute.

Find the rate of change of the height of the ladder at the time when the base is 20 feet from the base of the wall.

Draw a picture. Note that since height ($ y$) and distance from base ($ x$) are changing, we have to use variables for them. The ladder (slanted) will be the hypotenuse of a right triangle formed by the wall, the floor, and the ladder.
Define what we have and what we want when. We have: $ \displaystyle \frac{{dx}}{{dt}}=2$. We want: find $ \displaystyle \frac{{dy}}{{dt}}$ when $ x=20$.
Determine what equation we need. We need to relate the variables together; use the Pythagorean Theorem: $ {{x}^{2}}+{{y}^{2}}={{40}^{2}}$. Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time: $ \displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0$.
Plug in what we know at the time we know it, and solve for what we need. The tricky part here is that we need to use the Pythagorean Theorem again to get what $ y$ (height) is when $ x=20$ (distance from base of wall): $ {{20}^{2}}+{{y}^{2}}={{40}^{2}};\,\,\,y=\sqrt{{1200}}\approx 34.641$. Now plug everything in: $ \displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0;\,\,\,\,2\left( {20} \right)\left( 2 \right)+2\left( {34.641} \right)\frac{{dy}}{{dt}}=0$; $ \displaystyle \frac{{dy}}{{dt}}=-\frac{{2\left( {20} \right)\left( 2 \right)}}{{2\left( {34.641} \right)}}\approx -1.155\,\,\text{ft/min}$.
Make sure we’re answering the question. Yes, $ -1.155$ feet per minute is the rate the height of the ladder is changing. It makes sense that it is negative, since the ladder is slipping down the wall.

The width of a rectangle is increasing at a rate of 10 inches per second, and the area is increasing at a rate of 600 inches² per second.

When the width is 25 inches and the area is 1000 inches², how fast is the length of the rectangle changing?

Draw a picture. Note that since width ($ w$) and area ($ A$) are changing, we have to use variables for them. The length ($ l$) is also changing.
Define what we have and what we want when. We have: $ \displaystyle \frac{{dw}}{{dt}}=10$ and $ \displaystyle \frac{{dA}}{{dt}}=600$. We want: find $ \displaystyle \frac{{dl}}{{dt}}$ when $ w=25$ and $ A=1000$.
Determine what equation we need. We need to relate the variables together; use the area of a triangle: $ A=l\times w$. Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time: Use product rule: $ \displaystyle A=l\times w;\,\,\,\frac{{dA}}{{dt}}=l\cdot \frac{{dw}}{{dt}}+w\cdot \frac{{dl}}{{dt}}$.
Plug in what we know at the time we know it, and solve for what we need. The tricky part here is that we need to use the Area of a Rectangle again to get what $ l$ (length) is when $ w=25$ and $ A=1000$: $ A=l\times w;\,\,1000=l\times 25;\,\,l=40$. Now plug everything in: $ \displaystyle \frac{{dA}}{{dt}}=l\cdot \frac{{dw}}{{dt}}+w\cdot \frac{{dl}}{{dt}};\,\,\,600=40\cdot 10+25\cdot \frac{{dl}}{{dt}};\,\,\frac{{dl}}{{dt}}=8\,\,\text{in/min}$.
Make sure we’re answering the question. Yes, $ 8$ inches per minute is the rate the length is changing. It makes sense that it is positive, since the rectangle is getting larger.

A woman 5 feet tall walks at a rate of 6 feet per second away from a lamppost. When the woman is 6 feet from the lamppost, her shadow is 8 feet tall.

At what rate is the length of her shadow changing when she is 12 feet from the lamppost?

At what rate is the tip of her shadow changing when she is 12 feet from the lamppost?

Draw a picture. Remember that the woman’s shadow is on the ground in front of her and gets bigger as she moves away from the light; we’ll use $ y$ for this distance. The distance from the base of the lamppost to the woman is $ x$. We need variables for both of these distances, since they are changing.
Define what we have and what we want when. The rate the woman walks from the lamppost, $ \displaystyle \frac{{dx}}{{dt}}=6$. Now this is tricky: to get the rate the length of her shadow increases, we want $ \displaystyle \frac{{dy}}{{dt}}$, but if we want the rate the tip of her shadow increases, we want $ \displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}$. (This is because we need to measure the tip of the shadow in reference to the base of the lamppost). We also have the height of the woman (5 feet), and we know that when the woman is 6 feet from the lamppost $ (x=6),$ her shadow is 8 feet ($ y=8$). Since we will be using similar triangles to relate the triangles together before differentiating (no other way!), we will need the height $ h$ of the lamppost. We can get this using similar triangles when she is 6 feet away from the lamppost (see last triangle to the left): $ \displaystyle \frac{{\text{side of large }\Delta }}{{\text{side of small }\Delta }}=\frac{{\text{bottom of large }\Delta }}{{\text{bottom of small }\Delta }}:\,\,\,\,\,\,\frac{h}{5}=\frac{{14}}{8};\,\,h=8.75$.
Determine what equation we need. We need to relate $ x$ and $ y$ to what we know ($ h$ and woman’s height), which we can do using similar triangles again: $ \displaystyle \frac{{\text{8}\text{.75}}}{5}=\frac{{x+y}}{y},$ or (simplified), $ 3.75y=5x$. Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time. $ \displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}}$.
Plug in what we know at the time we know it, and solve for what we need. Now plug in: $ \displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}};\,\,\,\,3.75\frac{{dy}}{{dt}}=5\left( 6 \right);\,\,\,\frac{{dy}}{{dt}}=8$ feet per second. This is the rate of the length of the shadow.
To get the rate of the tip of the shadow, we need $ \displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}$, which is actually $ \displaystyle \frac{{dx}}{{dt}}+\frac{{dy}}{{dt}}$, by the sum rule (derivative of a sum is equal to the sum of the derivatives). The rate of the tip of the shadow is $ 6+8=14$ feet per second.

(Note that could have also defined a variable to represent $ x + y$, and do some subtraction in setting it up).

Here are problems that use more advanced Geometry:

Related Rates Problem	Steps and Solution
A snowball is melting at a rate of 1 cubic inch per minute. At what rate is the radius changing when the snowball has a radius of 2 inches? Note that $ \displaystyle {{V}_{{sphere}}}=\frac{4}{3}\pi {{r}^{3}}$.	Draw a picture. Note that since volume ($ V$) and radius ($ r$) are changing, we have to use variables for them. Define what we have and what we want when. We have: $ \displaystyle \frac{{dV}}{{dt}}=-1$ (since volume is decreasing). We want: find $ \displaystyle \frac{{dr}}{{dt}}$ when $ r=2$. Determine what equation we need. We need to relate the radius of a circle to its volume (sphere): $ \displaystyle {{V}_{{sphere}}}=\frac{4}{3}\pi {{r}^{3}}$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time. $ \displaystyle \require{cancel} \frac{{dV}}{{dt}}=\cancel{3}\cdot \frac{4}{{\cancel{3}}}\pi \,{{r}^{2}}\frac{{dr}}{{dt}}=4\pi \,{{r}^{2}}\frac{{dr}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need. $ \displaystyle \frac{{dV}}{{dt}}=4\pi {{r}^{2}}\frac{{dr}}{{dt}};\,\,$ $ \displaystyle -1=4\pi {{\left( 2 \right)}^{2}}\cdot \frac{{dr}}{{dt}};\,\,\,\frac{{dr}}{{dt}}=\frac{{-1}}{{16\pi }}\approx -.0199\,\,\text{in/min}$. Make sure we’re answering the question. Yes, $ -.0199$ inches per minute is the rate the radius is changing. It makes sense that it is negative, since radius is decreasing along with the volume.
Water is being poured at a rate of 3 cubic inches per second into a cylindrical can that is 20 inches in height and has a radius of 8 inches. How fast is the height of the water in the can changing when the height is 9 inches deep? Note that $ {{V}_{{cylinder}}}=\pi {{r}^{2}}h$.	Draw a picture. Note that since the radius and height of the actual can are not changing, we can use constants for them. Since the height ($ h$) and volume ($ V$) of the water inside are changing, we have to use variables for them. Define what we have and what we want when. We have: height of can $ =20$ (which we won’t need!), radius of can (and water) $ =8$, $ \displaystyle \frac{{dV}}{{dt}}=3$ (rate of volume of water in can). We want: find $ \displaystyle \frac{{dh}}{{dt}}$ (rate of height of water) when $ h=9$ (height of water; we also won’t need this since the rate of the height is constant!). Determine what equation we need: We need to relate the radius (a constant) and height of a cylinder to its volume: $ {{V}_{{cylinder}}}=\pi {{r}^{2}}h$. Differentiate with respect to time: $ \displaystyle \frac{{dV}}{{dt}}=\pi {{r}^{2}}\frac{{dh}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need: $ \displaystyle \frac{{dV}}{{dt}}=\pi {{r}^{2}}\frac{{dh}}{{dt}};\,\,\,\,\,3=\pi {{\left( 8 \right)}^{2}}\frac{{dh}}{{dt}};\,\,\,\frac{{dh}}{{dt}}\approx .015\,\,\text{in/sec}$. Make sure we’re answering the question: Yes, $ .015$ inches per second is the rate the height of the water is changing at any given time. It makes sense that it is positive, since the can is filling up. Note that the problem gave us extra information to confuse us!
A conical paper cup 4 inches across the top and 5 inches deep is full of a liquid. The cup springs a leak at the bottom and loses liquid at a rate of 2 cubic inches per minute. How fast is the liquid level changing at the instant when the liquid is exactly 3 inches deep? Note that $ \displaystyle {{V}_{{cone}}}=\frac{\pi }{3}{{r}^{2}}h$.	Draw a picture. Since the height ($ h$) and volume ($ V$) of the liquid inside are changing, we have to use variables for them. The radius ($ r$) of the liquid is also changing. Define what we have and what we want when. We have: $ \displaystyle \frac{{dV}}{{dt}}=-2$ (rate of volume of liquid in cup). We want: find $ \displaystyle \frac{{dh}}{{dt}}$ (rate of height of liquid) when $ h = 3$ (height of liquid). Determine what equation we need. We need to relate the radius and height of a cone to its volume: $ \displaystyle {{V}_{{cone}}}=\frac{\pi }{3}{{r}^{2}}h$. Make sure that all variables will differentiate to a rate we either have or need: yes! But we don’t have a value for a radius, so we need to relate the radius of the cone to the height of the cone using similar triangles: $ \displaystyle \frac{{\text{radius}}}{{\text{height}}}=\frac{2}{5}=\frac{r}{h};\,\,\,r=\frac{{2h}}{5}$. Now we have: $ \displaystyle {{V}_{{cone}}}=\frac{\pi }{3}{{r}^{2}}h=\frac{\pi }{3}{{\left( {\frac{{2h}}{5}} \right)}^{2}}h=\frac{{4\pi }}{{75}}{{h}^{3}}$. Differentiate with respect to time. $ \displaystyle \frac{{dV}}{{dt}}=3\cdot \frac{{4\pi }}{{75}}{{h}^{2}}\frac{{dh}}{{dt}}=\frac{{12\pi {{h}^{2}}}}{{75}}\frac{{dh}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need. Plug in: $ \displaystyle \frac{{dV}}{{dt}}=\frac{{12\pi {{h}^{2}}}}{{75}}\frac{{dh}}{{dt}};\,\,-2=\frac{{12\pi {{{\left( 3 \right)}}^{2}}}}{{75}}\frac{{dh}}{{dt}};\,\,\frac{{dh}}{{dt}}=-.442\,\,\text{in/min}$. Yes, $ -.442$ inches per minute is the rate the height of the liquid is changing at any given time. It makes sense that it is negative, since the volume is decreasing.

Here are a few that involve Right Triangle Trigonometry:

Related Rates Problems with Trig	Steps and Solution
A hot air balloon is rising straight up and is tracked by a range finder 400 feet from point of lift-off. When the range finder’s angle of elevation is $ \displaystyle \frac{\pi }{3}$, the angle is increasing at a rate of .15 radians per minute. How fast is the balloon rising at that time?	Draw a picture. Note that since height ($ h$) and the angle of elevation ($ \theta$) are changing, we have to use variables for them. Define what we have and what we want when. We have: $ \displaystyle \frac{{d\theta }}{{dt}}=.15$ (in radians). We want: find $ \displaystyle \frac{{dh}}{{dt}}$ when $ \displaystyle \theta =\frac{\pi }{3}$. We also know the distance on the ground from the range finder to the balloon is 400. Determine what equation we need. We need to relate the height and base of the right triangle to the angle of elevation; we can use trigonometry: $ \displaystyle \tan \theta =\frac{h}{{400}}$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time. $ \displaystyle \frac{{d\left( {\tan \theta } \right)}}{{dt}}=\frac{{d\left( {\frac{h}{{400}}} \right)}}{{dt}};\,\,\,{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}=\frac{1}{{400}}\frac{{dh}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need (calculator in radians): $ \displaystyle \begin{align}{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\left( {{{{\sec }}^{2}}\left( {\frac{\pi }{3}} \right)} \right)\left( {.15} \right)=\frac{1}{{400}}\frac{{dh}}{{dt}}\\{{2}^{2}}\left( {.15} \right)&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\frac{{dh}}{{dt}}=240\,\,\text{ft/min.}\end{align}$ $ \displaystyle (\text{Note that }\sec \left( {\frac{\pi }{3}} \right)=\frac{1}{{\cos \left( {\frac{\pi }{3}} \right)}}=\frac{1}{{\frac{1}{2}}}=2)$. Make sure we’re answering the question. Yes, $ 240$ feet per minute is the rate the balloon is rising. It makes sense that it is positive since the height is increasing along with the angle size.
Here’s one similar, but we have to find the rate of an angle: A balloon is rising straight up at a rate of 3 feet per second. A woman letting go of the balloon runs 12 feet way and observes its rising. How fast is the angle of observation changing (in radians per second) when the balloon is 20 feet from the ground?	Draw a picture. Note that since height ($ h$) and the angle of elevation ($ \theta$) are changing, we have to use variables for them. Define what we have and what we want when. We have: $ \displaystyle {{dh} \over {dt}} = \,\,3$. We want: find $ \displaystyle {{d\theta } \over {dt}}$ when $ h = 20$. We also know the distance on the ground from where the balloon took off to the woman is 12. Determine what equation we need. We need to relate the height and base of the right triangle to the angle of elevation; we can use trigonometry: $ \displaystyle \tan \theta =\frac{{\text{Opp}}}{{\text{Adj}}}=\frac{h}{{12}}$. Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time. $ \displaystyle \frac{{d\left( {\tan \theta } \right)}}{{dt}}=\frac{{d\left( {\frac{h}{{12}}} \right)}}{{dt}};\,\,\,\,\,{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}=\frac{1}{{12}}\frac{{dh}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need. The trick here is to get $ \theta $ when $ \displaystyle h=20$. There are two ways to do this: 1. Use an inverse trig calculation with the calculator (radians): $ \displaystyle \tan \left( \theta \right)=\frac{{20}}{{12}};\,\,\theta ={{\tan }^{{-1}}}\left( {\frac{{20}}{{12}}} \right)\approx 1.0304$ $ \displaystyle {{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}=\frac{1}{{12}}\frac{{dh}}{{dt}}\,$ $ \displaystyle {{\sec }^{2}}\left( {1.0304} \right)\frac{{d\theta }}{{dt}}=\frac{1}{{12}}\left( 3 \right);\,\,\,\frac{{d\theta }}{{dt}}\approx \frac{{.25}}{{3.778}}\approx .0662\,\,\text{rad/sec}$ 2. Use the Pythagorean Theorem and right triangle trigonometry: $ {{12}^{2}}+{{20}^{2}}={{\left( {\text{hypotenuse}} \right)}^{2}};\,\,\text{hypotenuse}\approx 23.324$ $ \displaystyle {{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}=\frac{1}{{12}}\frac{{dh}}{{dt}};\,\,\,\,\,\left( {\sec \left( \theta \right)=\frac{{\text{Hyp}}}{{\text{Adj}}}} \right)$ $ \displaystyle \,{{\left( {\frac{{23.324}}{{12}}} \right)}^{2}}\frac{{d\theta }}{{dt}}=\frac{1}{{12}}\left( 3 \right);\,\,\,\frac{{d\theta }}{{dt}}\approx \frac{{.25}}{{3.778}}\approx .0662\,\,\text{rad/sec}$ Make sure we’re answering the question. Yes, .0662 radians per second is the rate the angle of observation is increasing. It makes sense that it is positive since it increases when the height increases.
Two women are 35 feet apart. One of them starts walking south at a rate so that the angle below in the diagram is changing at a constant rate of 0.02 radians per minute. At what rate is the distance between the two women changing when $ \theta =.5$ radians?	Draw a picture. Note that since $ \theta $ and the distance between the women are changing, we have to use variables for them. Let $ x=$ the distance between the two women. Define what we have and what we want when. We have: $ \displaystyle \frac{{d\theta }}{{dt}}=\,\,.02$ (in radians). We want: find $ \displaystyle \frac{{dx}}{{dt}}$ when $ \displaystyle \theta =.5$. We also know the original distance between the two women is 35. Determine what equation we need. We need to relate the given angle to the sides of the triangle; we can use trigonometry: .$ \displaystyle \cos \theta =\frac{{35}}{x}$. As we’ll see below, we’ll also need to get $ x$ and relate it back to $ \theta $ when $ \theta =.5$; we’ll use this same equation: $ \displaystyle \cos \left( {.5} \right)=\frac{{35}}{x};\,\,x\approx 39.882$. Differentiate with respect to time. $ \displaystyle \frac{{d\left( {\cos \theta } \right)}}{{dt}}=\frac{{d\left( {35{{x}^{{-1}}}} \right)}}{{dt}};\,\,\,\,\,-\sin \theta \frac{{d\theta }}{{dt}}=-35{{x}^{{-2}}}\frac{{dx}}{{dt}}$. Plug in what we know at the time we know it, and solve for what we need (calculator in radians). $ \displaystyle \begin{align}\,-\sin \theta \frac{{d\theta }}{{dt}}&=-35{{x}^{{-2}}}\frac{{dx}}{{dt}};\,\,\,\,\,\left( {-\sin \left( {.5} \right)} \right)\left( {.02} \right)=-35{{\left( {39.882} \right)}^{{-2}}}\frac{{dx}}{{dt}}\,\\\frac{{dx}}{{dt}}&=\frac{{\left( {-\sin \left( {.5} \right)} \right)\left( {.02} \right){{{\left( {39.882} \right)}}^{2}}}}{{-35}}\approx .436\,\,\text{ft/min.}\end{align}$ Make sure we’re answering the question. Yes, $ .436$ feet per minute is the rate the distance between the two women is changing. It makes sense that it is positive since the angle between them is increasing.

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