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AP Calculus AB : Applications of Derivatives

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Example Question #21 : Interpretation Of The Derivative As A Rate Of Change

The amount of bacteria in a petri dish at any time, t, is modelled by:

$b(t)=\frac{2}{5}e^{\frac{t}{3}}$

What is the rate of change of the amount of bacteria at t=30?

Possible Answers:

$\frac{2}{15}e^{10}$

$e^{10}$

$\frac{2}{15}e^{5}$

$\frac{2}{15}e^{30}$

Correct answer:

$\frac{2}{15}e^{10}$

Explanation:

The rate of change of the amount of bacteria in the dish at any time, t, is given by the first derivative:

$b'(t)=\frac{2}{15}e^{\frac{t}{3}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d}}{\mathrm{d} x}ax=a$

Evaluating the first derivative at t=30, we get

$\frac{2}{15}e^{\frac{30}{3}}=\frac{2}{15}e^{10}$

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Example Question #51 : Applications Of Derivatives

A family of eight wants to track how much cereal they eat. Determine how fast is the cereal in the box decreasing if the amount of cereal at any time is given by

c(t)=104-24t

Possible Answers:

-24t

There is not enough information given

Correct answer:

Explanation:

To find the rate of change of the amount of cereal in the box at any time, we must take the derivative of the function:

c'(t)=-24

which was found using the rule

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

(We can tell, therefore, that the cereal is decreasing at a constant rate.)

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Example Question #11 : Calculus 3

Let $f(x)=x^2-\frac{1}{1-x^2}$ . Which of the following gives the equation of the line normal to $f(x)$ when x=2 ?

Possible Answers:

$32x+9y=103$

$27x+111y=535$

$27x + 96y = 470$

$32x-9y=25$

$192x + 54y = 735$

Correct answer:

$27x + 96y = 470$

Explanation:

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at x=2 . In order to find the tangent line, we will need to evaluate the derivative of f(x) at x=2 .

$f(x)=x^2-\frac{1}{1-x^2}=x^2-(1-x^2)^{-1}$

$f'(x)=2x-(-1)(1-x^2)^{-2}(-2x)$

$f'(x)=2x-2x(1-x^2)^{-2}$

$f'(2)=2(2)-2(2)(1-2^2)^{-2}$

$f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at x=2 is $\frac{32}{9}$ . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

$\left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at x=2 , it will pass through the point $\left ( 2,f(2) \right )$ . Be careful to use the original equation for , not its derivative.

$f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $-\frac{9}{32}$ and passes through the piont $\left ( 2,\frac{13}{3} \right )$ . We can now use point-slope form to find the equation of the normal line.

$y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by 32(3)=96 .

$96y-416=-27(x-2)$

$27x + 96y = 470$

The answer is $27x + 96y = 470$ .

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Example Question #1 : Implicit Differentiation

Use implicit differentiation to find $\frac{\mathrm{d}y }{\mathrm{d} x}$ given

sin(x^2)+cos(y^2)=1

Possible Answers:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{ysin(y^2)}{xcos(x^2)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{cos(2x)}{sin(2y)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2xcos(x^2)}{1+2ycos(y^2)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{xcos(x^2)}{ysin(y^2)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{xsin(x^2)}{ycos(y^2)}$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{xcos(x^2)}{ysin(y^2)}$

Explanation:

We simply differentiate both sides of the equation

sin(x^2)+cos(y^2)=1

2xcos(x^2)dx-2ysin(y^2)dy=0 (Don't forget the chain rule)

Now we solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$

xcos(x^2)dx=ysin(y^2)dy

$xcos(x^2)=\frac{ysin(y^2)dy}{dx}$

$\frac{xcos(x^2)}{ysin(y^2)}=\frac{\mathrm{d}y }{\mathrm{d} x}$

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Example Question #1 : Implicit Differentiation

Find $\frac{dy}{dx}.$

x^2+3y^2=9

Possible Answers:

$\frac{dy}{dx}=\frac{-x}{3y}$

$\frac{dy}{dx}=\frac{x}{3y}$

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{dy}{dx}=\frac{-x^2}{3y^2}$

$\frac{dy}{dx}=\frac{x}{y}$

Correct answer:

$\frac{dy}{dx}=\frac{-x}{3y}$

Explanation:

Implicit differentiation is very similar to normal differentiation, but every time we take a derivative with respect t , we need to multiply the result by $\frac{dy}{dx}.$ We also differentiate the entire equation from left to right, including any numbers. Then, we solve for that $\frac{dy}{dx}$ for our final answer.

x^2+3y^2=9

$2x+6y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-2x}{6y}=\frac{-x}{3y}.$

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Example Question #3 : Implicit Differentiation

Find $\frac{dy}{dx}.$

y^2=4sinx+3x^2

Possible Answers:

$\frac{dy}{dx}=\frac{-2cosx+3x}{y^2}$

$\frac{dy}{dx}=\frac{2sinx+3x}{y}$

$\frac{dy}{dx}=\frac{2cosx+3x}{y^2}$

$\frac{dy}{dx}=-2cosx+3x$

$\frac{dy}{dx}=\frac{2cosx+3x}{y}$

Correct answer:

$\frac{dy}{dx}=\frac{2cosx+3x}{y}$

Explanation:

Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to , we need to implicitly write $\frac{dy}{dx}$ after it. Hence, the name of this method. Then, we solve for $\frac{dy}{dx}.$

$(y^2=4sinx+3x^2)'=2y\frac{dy}{dx}=4cosx+6x.$

$\frac{dy}{dx}=\frac{4cosx+6x}{2y}=\frac{2cosx+3x}{y}.$

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Example Question #4 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function with respect to x

$y=xy^2+2\tan(y)$

Possible Answers:

$y'=\frac{y}{1-2xy-2\sec^2{y}}$

$y'=\frac{y^2}{1-2xy-2\sec^2{x}}$

$y'=\frac{y^2}{1+2xy-2\sec^2{y}}$

$y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

Correct answer:

$y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

Explanation:

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as .

$y'=\frac{\mathrm{d} }{\mathrm{d} x}(xy^2)+\frac{\mathrm{d} }{\mathrm{d} x}(2\tan{y})$

$y'=y^2+y'(2xy)+2y'\sec^2(y)$

Bringing the terms with to one side and factoring it out, we get

$y'(1-2xy-2\sec^2y)=y^2$

$y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

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Example Question #5 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function

y=x^3y+y^2

Possible Answers:

$y'=\frac{4x^2y}{1-x^2-2y}$

$y'=\frac{3x^2y}{1-x^3-2y}$

$y'=\frac{3x^2y}{1-x^2-y}$

$y'=\frac{3x^2y}{1+x^3+2y}$

Correct answer:

$y'=\frac{3x^2y}{1-x^3-2y}$

Explanation:

$y'=\frac{\mathrm{d} }{\mathrm{d} x}(x^3y)+\frac{\mathrm{d} }{\mathrm{d} x}(y^2)$

y'=y'x^3+3x^2y+2yy'

Bringing the terms with to one side and factoring it out, we get

y'(1-x^3-2y)=3x^2y

$y'=\frac{3x^2y}{1-x^3-2y}$

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Example Question #1 : Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

y+x^2+e^xy^2=2y^2

Possible Answers:

$\frac{-2x+e^xy^2}{1+2ye^x-4y}$

$\frac{-2x-e^xy^2}{2ye^x-4y}$

$\frac{-2x-e^xy^2}{1+2ye^x-4y}$

$\frac{-2x-e^xy^2}{1-2ye^x+4y}$

$\frac{2x-e^xy^2}{1+2ye^x-4y}$

Correct answer:

$\frac{-2x-e^xy^2}{1+2ye^x-4y}$

Explanation:

To find $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we must take the derivative of both sides of the equation with respect to x. When we do this, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}+2x+e^xy^2+2ye^x\frac{\mathrm{d}y }{\mathrm{d} x}=4y\frac{\mathrm{d} y}{\mathrm{d} x}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

Note that for every derivative of a function of y with respect to x, the chain rule was used, which accounts for $\frac{\mathrm{d} y}{\mathrm{d} x}$ appearing.

Algebraic simplification gets us our final answer,

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2x-e^xy^2}{1+2ye^x-4y}$

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Example Question #1 : Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

$\cos(xy)=2x$

Possible Answers:

$-\frac{2+\sin(xy)-y}{x}$

$-\frac{2+x\sin(xy)}{y\sin(xy)}$

$\frac{2+y\sin(xy)}{x\sin(xy)}$

$-\frac{2+y\sin(xy)}{x\sin(xy)}$

$\frac{2+x\sin(xy)}{y\sin(xy)}$

Correct answer:

$-\frac{2+y\sin(xy)}{x\sin(xy)}$

Explanation:

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

$-\sin(xy)(y+x\frac{\mathrm{d} y}{\mathrm{d} x}) =2$

The following derivative rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the chain rule was used for both the cosine function (which contains an inner product of two functions), and for the derivative of y, whose derivative with respect to x we want to solve for.

Solving, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2+y\sin(xy)}{x\sin(xy)}$

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AP Calculus AB : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Interpretation Of The Derivative As A Rate Of Change

Example Question #51 : Applications Of Derivatives

Example Question #11 : Calculus 3

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #3 : Implicit Differentiation

Example Question #4 : Implicit Differentiation

Example Question #5 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

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