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Calculus AB : Contextual Applications of Derivatives

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Example Questions

Example Question #21 : Contextual Applications Of Derivatives

Find the velocity function from an acceleration function given by

a(t)=4t+5

and the condition v(0)=10

Possible Answers:

2t^2+5t+C

2t^2+5t

2t^2+15

2t^2+5t+10

Correct answer:

2t^2+5t+10

Explanation:

Acceleration is the rate of change of velocity, so we must integrate the acceleration function to find the velocity function:

$v(t)=\int a(t)dt=\int (4t+5 ) dt=2t^2+5t+C$

The integration was performed using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int a dx=ax+C$

To find the integration constant C, we must use the initial condition given:

v(0)=10=2(0)^2+5(0)+C

C=10

Our final answer is

v(t)=2t^2+5t+10

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Example Question #321 : Calculus Ab

The velocity of a particle is given by v(t). Find the function which models the particle's acceleration.

v(t)=3t^3-4t^2-5t+16

Possible Answers:

a(t)=9t^2-8t-5

a(t)=9t^3+8t^2-5

a(t)=9t^2-8t-17

a(t)=6t^2-8t+5

Correct answer:

a(t)=9t^2-8t-5

Explanation:

The velocity of a particle is given by v(t). Find the function which models the particle's acceleration.

v(t)=3t^3-4t^2-5t+16

To find the acceleration from a velocity function, simply take the derivative.

In this case, we are given v(t), and we need to find v'(t) because v'(t)=a(t).

To find v'(t), we need to use the power rule.

For each term, simply multiply by the exponent, and then subtract one from the exponent. Constant terms will drop out, linear terms will become constants, and so on.

v(t)=3t^3-4t^2-5t+16

$v'(t)=a(t)=(3)3t^{3-1}-(2)4t^{2-1}-5=9t^2-8t-5$

So, our answer is:

a(t)=9t^2-8t-5

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Example Question #11 : Calculate Position, Velocity, And Acceleration

The velocity of a particle is given by v(t) . Find the particle's acceleration when t=3 .

v(t)=3t^3-4t^2-5t+16

Possible Answers:

a=53

a=27

a=-3

a=52

Correct answer:

a=52

Explanation:

The velocity of a particle is given by v(t) . Find the particle's acceleration when t=3 .

v(t)=3t^3-4t^2-5t+16

To find the acceleration from a velocity function, simply take the derivative.

In this case, we are given v(t) , and we need to find v'(t) because v'(t)=a(t) .

To find v'(t) , we need to use the power rule.

For each term, simply multiply by the exponent, and then subtract one from the exponent. Constant terms will drop out, linear terms will become constants, and so on.

v(t)=3t^3-4t^2-5t+16

$v'(t)=a(t)=(3)3t^{3-1}-(2)4t^{2-1}-5=9t^2-8t-5$

So, our acceleration function is:

a(t)=9t^2-8t-5

Now, plug in for and solve.

a(3)=9(3)^2-8(3)-5=81-24-5=52

So, our answer is .

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Example Question #1 : Calculate Position, Velocity, And Acceleration

The position of a car is given by the following function:

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Example Question #1 : Contextual Applications Of Derivatives

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=e^{\pi }$

$f''(x)= e^{x}$

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

$f'(x)=e^{\pi x }$

$f''(x)=e^{\pi x}$

$f'(x)=\pi$

$f''(x)=\0$

Correct answer:

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

Explanation:

In order to solve for the first and second derivatives, we must use the chain rule.

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derivative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[e^u]$

$\frac{dy}{du}=e^u$

And differentiating we use the power rule which states

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

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

$f'(x)=\pi e^{\pi x}$

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

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

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

$f''(x)=\pi^2e^{\pi x}$

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Example Question #1 : Calculate Position, Velocity, And Acceleration

Find the velocity function of the particle if its position is given by the following function:

$s(t)=e^t\tan(t)+t^5+3$

Possible Answers:

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

Correct answer:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Explanation:

The velocity function is given by the first derivative of the position function:

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Example Question #1 : Calculate Position, Velocity, And Acceleration

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=cot(x)

f''(x)=csc^2(x)

f'(x)=sec^2(x)

f''(x)=sec^3(x)

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

f'(x)=csc(x)

f''(x)=-sec(x)

Correct answer:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

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Example Question #2 : Calculate Position, Velocity, And Acceleration

Given the velocity function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

a(x) .

Possible Answers:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(2x+2)^{d}(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(x+1)^{d}\sin (x+1)^{10}$

$a(x)=(x^2+2x+1)^{d}\sin (x^2+2x+1)^{10}$

$a(x)=(x^2+2x+1)^{2d}\sin (x^2+2x+1)^{10}$

Correct answer:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Explanation:

We can find the acceleration function a(x) from the velocity function by taking the derivative:

a(x)=v'(x)

We can view the function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

as the composition of the following functions

$g(x)=\int_{0}^{x} z^d\sin(z^5) dz$

h(x)=x^2+2x+1=(x+1)^2

so that v(x)=g(h(x)) . This means we use the chain rule

[g(h(x))]'=g'(h(x))h'(x)

to find the derivative. We have $g'(x)=x^d\sin x^5$ and h'(x)=2x+2 , so we have

$a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)$

$=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

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Example Question #2 : Calculate Position, Velocity, And Acceleration

The position of an object is given by the equation $y = \frac{1}{3} t^3 - \frac{1}{2} t^2 -2t + 4$ . What is its acceleration at t=2 ?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

y ' = t^2 - t -2

y'' = 2t - 1

Now plug in 2 for t:

y'' = 2(2) - 1 = 4- 1 = 3

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Example Question #1 : Contextual Applications Of Derivatives

The equation $y= \sin (2t)$ models the position of an object after t seconds. What is its speed after $\pi$ seconds?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed.

$y' = \cos (2t ) \cdot 2$

Plug in $\pi$ for t:

$2 \cos (2 \pi ) = 2\cdot 1 = 2$

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Calculus AB : Contextual Applications of Derivatives

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #21 : Contextual Applications Of Derivatives

Example Question #321 : Calculus Ab

Example Question #11 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Contextual Applications Of Derivatives

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #2 : Calculate Position, Velocity, And Acceleration

Example Question #2 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Contextual Applications Of Derivatives

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