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AP Calculus AB : Applications of antidifferentiation

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Example Question #6 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Find the velocity function given the following information:

The acceleration function is $a(t)=3t+\cos(t)$ ;

v(0)=3

Possible Answers:

$\frac{t^2}{2}+\sin(t)+3$

$\frac{3t^2}{2}-\sin(t)+3$

$\frac{3t^2}{2}+\sin(t)+3$

$\frac{3t^2}{2}+\sin(t)+C$

Correct answer:

$\frac{3t^2}{2}+\sin(t)+3$

Explanation:

To find the velocity function, we integrate the acceleration function (the acceleration is the antiderivative of the velocity):

$v(t)=\int a(t)dt=\int [3t +\cos(t)]dt =\frac{ 3t^2}{2}+\sin(t)+C$

The rules of integration used were

$\int x^n dx = \frac{x^{n+1}}{n+1}+C$ , $\int \cos(t)dt= \sin(t)+C$

To solve for the integration constant, we plug in the given initial condition:

v(0)=3=0+0+C

3=C

Our final answer is

$v(t)=\frac{3t^2}{2}+\sin(t)+3$

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Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

What is the position function if the initial position is 0 and the velocity function is given by v(t)=10t+1 ?

Possible Answers:

5t^2+1

$\frac{5t^2}{2}+t$

5t^2+t

Correct answer:

5t^2+t

Explanation:

To find the position function, we must integrate the velocity function, as velocity is the antiderivative of position:

$s(t)=\int v(t) dt = \int (10t+1)dt = 5t^2+t+C$

The following rule of integration was used:

$\int x^n dx = \frac{x^{n+1}}{n+1}+C$

Finally, we use the initial condition to solve for the integration constant:

s(0)=0=0+0+C

C=0

Our final answer is

s(t)=5t^2+t

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Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

A particle at the origin has an initial velocity of . If its acceleration is given by a = -24t^2 + 6t - 8 , find the position of the particle after 1 second.

Possible Answers:

24m

29m

32m

18m

27m

Correct answer:

27m

Explanation:

In this problem, letting s(t) denote the position of the particle and v(t) denote the velocity, we know that s''(t) = v'(t) = a(t) . Integrating and working backwards we have,

$v(t) = \int a(t)dt = \int (-24t^2 + 6t - 8)dt = -8t^3 + 3t^2 -8t + C$

Plugging in our initial condition, v(0) = 32 , we see immediately that C = 32 .

Repeating the process again for s(t) , we find that

$s(t) = \int v(t)dt = \int (-8t^3 + 3t^2 -8t +32)dt = -2t^4 + t^3 - 4t^2 + 32t + C$

Plugging in our initial condition, s(0) = 0 (we started at the origin) we see that C = 0 . This gives us a final equation

s(t) = -2t^4 + t^3 - 4t^2 + 32t . The problem asks for s(1) which is simply s(t) = -2(1)^4 + (1)^3 - 4(1)^2 + 32(1) = -4 + 1 - 4 + 32 = 27

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Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Find the integral which satisfies the specific conditions of $(\frac{\pi}{2}, 4)$

$\int4sin(t)-12cos(t)+3t^2dt$

Possible Answers:

y=4cos(t)+12sin(t)+t^3

y=-16tan(t)+t^3+12.12

y=-4tcos(t)-12tsin(t)+t^3

y=-4cos(t)-12sin(t)+t^3+12.12

Correct answer:

y=-4cos(t)-12sin(t)+t^3+12.12

Explanation:

Find the integral which satisfies the specific conditions of $(\frac{\pi}{2}, 4)$

$\int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called anti-derivatives. This means that we can calculate integrals by reversing our integration rules.

Furthermore, to find the specific answer using initial conditions, we need to find our "c" at the end.

Thus, we can have the following rules.

$\int sin(t)dt = -cos(t)+c$

$\int cos(t)dt = sin(t)+c$

$\int t^ndt=\frac{t^{n+1}}{n+1}+c$

Using these rules, we can find our answer:

$\int4sin(t)-12cos(t)+3t^2dt$

Will become:

$-4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c$

And so our anti-derivative is:

-4cos(t)-12sin(t)+t^3+c

Now, let's find c. First set our above expression equal to y

y= -4cos(t)-12sin(t)+t^3+c

Next, plug in for y and t. Then solve for c

$4= -4cos(\frac{\pi}{2})-12sin(\frac{\pi}{2})+(\frac{\pi}{2})^3+c$

Looks a bit messy, but we can clean it up to get:

$4= 0-12+3.88+c\rightarrow 4=-8.12+c\rightarrow c=12.12$

Now, to solve, simply replace c with 12.12

y=-4cos(t)-12sin(t)+t^3+12.12

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AP Calculus AB : Applications of antidifferentiation

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #6 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

All AP Calculus AB Resources

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