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AP Calculus AB : Derivatives

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Example Questions

Example Question #581 : Derivatives

A fluid cools with time according to the equation

$\frac{\mathrm{d} T}{\mathrm{d} t}=hA(T-30)$

What is the cooling rate at a temperature of 29?

Possible Answers:

-hA

Not enough information given

Correct answer:

-hA

Explanation:

The equation given is the rate of change of temperature itself, so to find the cooling rate at a temperature (T) of 29, we simply plug this into the rate function:

$\frac{\mathrm{d} T}{\mathrm{d} t}=hA(-1)=-hA$

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Example Question #44 : Applications Of Derivatives

Banana farmers notice a fungi is destroying bananas. They will use pesticides when the bananas die at a rate of 31 per day. What time will this occur if the amount of dead bananas at any time is given by

b(t)=10t^2+t+5 , where t is time in days.

Possible Answers:

None of the other answers

$\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

In order to find the time at which the banana death rate is 31, we must take the derivative of the function for dead bananas, whose rate of change - the death rate - is given by:

b'(t)=20t+1

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

To find the time at which the banana death rate is equal to 31, we set the derivative function equal to 31 and solve for t:

31=20t+1

30=20t

$\frac{3}{2}=t$

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Example Question #45 : Applications Of Derivatives

Find the velocity at t=3 given the position function

$s(t)=10e^{t^2}$

Possible Answers:

60e^6

60e^9

20e^9

10e^9

Correct answer:

60e^9

Explanation:

The velocity is determined by taking the derivative of the position function, as velocity is the rate of change of position:

$v(t)=s'(t)=20te^{t^2}$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we simply plug in the given t value into the velocity function:

v(3)=60e^9

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Example Question #46 : Applications Of Derivatives

Find the time at which the rate of change of the function is equal to 3:

$f(t)=\frac{t^3}{3}-t^2, t\geq0$

Possible Answers:

t=-1

t=3

t=0

t=3, -1

Correct answer:

t=3

Explanation:

To find the rate of change of the function at any instant, we must take its first derivative:

f'(t)=t^2-2t

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, to find the value of t at which the rate of change is equal to 3, we use algebra:

3=t^2-2t

t^2-2t-3=0

t=3, -1

The problem statement tells us that t cannot be negative, so our only answer is

t=3

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Example Question #582 : Derivatives

Mass accumulates in a bag according to the function:

m(t)=2t+5

What is the rate of change of mass in the bag at t=1?

Possible Answers:

None of the other answers

Correct answer:

Explanation:

The rate of change of mass in the bag at any instant is given by the first derivative of the function:

m'(t)=2

which was found using the following rule:

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

Notice that the rate of change is a constant value, not dependent on t. (You can compare the original equation to a line which has a slope of 2; the rate of change of the line is always 2, regardless of the t value you are at!)

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Example Question #51 : Applications Of Derivatives

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:

$e^{10}$

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

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

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

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 #64 : Derivatives Of Functions

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:

$27x+111y=535$

$27x + 96y = 470$

$32x+9y=103$

$192x + 54y = 735$

$32x-9y=25$

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:

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

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

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

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

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

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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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #581 : Derivatives

Example Question #44 : Applications Of Derivatives

Example Question #45 : Applications Of Derivatives

Example Question #46 : Applications Of Derivatives

Example Question #582 : Derivatives

Example Question #51 : Applications Of Derivatives

Example Question #51 : Applications Of Derivatives

Example Question #64 : Derivatives Of Functions

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

All AP Calculus AB Resources

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