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AP Calculus AB : Applications of antidifferentiation

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Example Questions

Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{4y}{x^2}, x>0$

and at x=1, y=2

Possible Answers:

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

$y=Ce^{-\frac{4}{x}}$

$y=2e^{-\frac{4}{x}}+C$

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

y=2

Correct answer:

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

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

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

Next, we integrate both sides:

$\ln \left | y\right | = -\frac{1}{x}+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}= \ln \left | x\right |+C$ , $\int x^n dx = \frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y:

$y=e^{-\frac{1}{x}+C}$

Using the properties of exponents, we can rearrange the integration constant:

$y=e^{-\frac{1}{x}+C} = e^{-\frac{1}{x}}\cdot e^C = Ce^{-\frac{1}{x}}$

(The exponential of the constant is itself a constant.)

Using the given condition, we can solve for C:

$2=Ce^{-4}$

C=2e^4

Our final answer is

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

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Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

The rate of a chemical reaction is given by the following differential equation:

$\frac{-d[A]}{dt} = k[A]^{2}$ ,

where $\left [ A \right ]$ is the concentration of compound at a given time, . Which one of the following equations describes $\frac{1}{\left [ A \right ]}$ as a function of time? Let $\left [ A \right ]_{0}$ be the concentration of compound when t=0 .

Possible Answers:

$\frac{1}{[A_{0}]} + kt = \frac{1}{[A]}$

$[A_{0}]e^{kt}= \frac{1}{[A]}$

$[A_{0}]kt= \frac{1}{[A]}$

$[A_{0}]e^{-kt}= \frac{1}{[A]}$

$\frac{1}{[A_{0}]} - kt = \frac{1}{[A]}$

Correct answer:

$\frac{1}{[A_{0}]} + kt = \frac{1}{[A]}$

Explanation:

To begin with, the differential equation needs to be rearranged so that each variable is one side of the equation:

$\frac{-d[A]}{dt} = k[A]^{2} \Rightarrow \frac{d[A]}{[A]^{2}} = -kdt$ .

Then, integrate each side of the rate law, bearing in mind that $\left [ A \right ]$ will range from $\left [ A \right ]_{0}$ to $\left [ A \right ]$ , and time will range from to :

$\int_{[A]_{0}}^{[A]}\frac{d[A]}{[A]^{2}} = \int_{0}^{t} -kdt$

After integrating each side, the equation becomes:

$\frac{-1}{[A]} = -kt$ .

The left side has to be evaluated from $\left [ A \right ]_{0}$ to $\left [ A \right ]$ , and the right side is evaluated from to :

$\frac{-1}{[A]} - ( \frac{-1}{[A_{0}]}) = -kt - (-k(0))$ . This becomes:

$\frac{1}{[A_{0}]} - \frac{1}{[A]} = -kt$ .

Finally, rearranging gives:

$\frac{1}{[A_{0}]} + kt = \frac{1}{[A]}$

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Example Question #21 : Applications Of Antidifferentiation

Given that y' = (2y + 2)(x -2) and y(0) = 1 , solve for y(t) . What is the value of ?

Possible Answers:

2e^2 - 1

2e - 1

Correct answer:

Explanation:

This is a separable differential equation. The simplest way to approach this is to turn into $\frac{\mathrm{d} y}{\mathrm{d} x}$ , and then by abusing the notation, "multiplying by dx" on both sides.

dy = (2y+2)(x - 2)dx

We then group all the y terms with dy, and all the x terms with dx.

$\frac{1}{2y+2}dy = (x - 2)dx$

Integrating both sides, we find

$\int \frac{1}{2y+2}dy = \frac{\ln|2y+2|}{2} =\int (x - 2)dx = \frac{x^2}{2} - 2x + C$

Here, the first integral is found by using substitution of variables, setting u = 2y +2 . In addition, we have chosen to only put a +C on the second integral, as if we put it on both, we would just combine them in any case.

To solve for y, we multiply both sides by two and raise e to both sides to get rid of the natural logarithm.

$|2y+2| = e^{x^2 - 4x + C}$

(Note, C was multiplied by two, but it's still just an arbitrary constant. If you prefer, you may call the new C value C_1 .)

Now we drop our absolute value signs, and note that we can take out a factor of e^C and stick in front of the right hand side.

$2y+2 =\pm e^Ce^{x^2 - 4x}$

As $\pm e^C$ is just another arbitrary constant, we can relabel this as C, or C_2 if you prefer. Solving for y gets us

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

Next, we plug in our initial condition to solve for C.

$1 =\frac{C - 2}{2}$ ; C = 4

Leaving us with a final equation of

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

Plugging in x = 4, we have a final answer,

$y = \frac{4e^{16 - 16} - 2}{2} = \frac{4 - 2}{2} = 1$

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Example Question #21 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

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

given the condition at

$x=\pi, y=2$

Possible Answers:

$y=e^{\sin(x)}$

y=x

$y=Ce^{\sin(x)}$

$y=2e^{\sin(x)}$

Correct answer:

$y=2e^{\sin(x)}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$\frac{dy}{y}=\cos(x)dx$

Next, we integrate both sides:

$\int \frac{dy}{y}=\ln\left | y\right |+C$ ; $\int \cos(x)dx=\sin(x)+C$

The integrals were solved using the identical rules.

The two constants of integration are now combined to make a single one:

$\ln \left | y\right |= \sin(x)+C$

Now, exponentiate both sides of the equation to solve for y, and use the properties of exponents to rearrange C:

$y=e^{\sin(x)+C}=e^{\sin(x)}\cdot e^C=Ce^{\sin(x)}$

Finally, we solve for the integration constant using the given condition:

$2=Ce^{\sin(\pi)}=Ce^0=C$

Our final answer is

$y=2e^{\sin(x)}$

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Example Question #21 : Applications Of Antidifferentiation

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

h(t)=-15t^2+30t feet.

At what time will the velocity equal zero?

Possible Answers:

1.5s

Correct answer:

Explanation:

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t²+ 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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

A particle is moving in a straight path with a constant initial velocity. The particle is then subjected to a force causing a time-dependent acceleration given as a function of time:

a(t)=(a+b)t

After 10 seconds, the particle has a velocity equal to meters-per-second. Find the initial velocity in terms of the constants , and

Units are all in S.I. (meters, seconds, meters-per-second, etc.)

Possible Answers:

v_o =k-50(a+b)

v_o =b-5(a+k)

$v_o =\frac{b}{a}+k$

$v_o =\frac{k}{a+b}-50a$

$v_o =\frac{k}{a}+b$

Correct answer:

v_o =k-50(a+b)

Explanation:

a(t)=(a+b)t

Begin by finding the velocity function by integrating the acceleration function.

$v(t)=\int a(t)dt = \int (a+b)tdt=\frac{1}{2}(a+b)t^2+v_o$

We use v_o as the constant of integration since the function v(t) is a velocity and at initially, at t = 0 , the velocity is equals the constant of integration.

$v(t)=\frac{1}{2}(a+b)t^2+v_o$

At t = 10 seconds we are told the velocity is equal to .
$v(t=10)=k\: \: \: \: \: \:\rightarrow\: \: \: \: \: \:\frac{1}{2}(a+b)100+v_o =k$

50(a+b)+v_o =k

v_o =k-50(a+b)

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

Find the average value of the function f(x)=cosx on the interval $[0, \pi].$

Possible Answers:

$\frac{2}{\pi}$

$\frac{1}{\pi}$

$\frac{-1}{\pi}$

Correct answer:

Explanation:

The average value of a function on a given interval [a,b] is given by the following function:

$\frac{1}{b-a}\int_{a}^{b}f(x)dx.$

Now, let's simply input our values and function in:

$\frac{1}{\pi-0}\int_{0}^{\pi}-cosxdx=\frac{1}{\pi}[sin(\pi)-sin(0)]=\frac{0}{\pi}*0=0.$

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

Determine the position function for a particle whose velocity is given by the equation

v(t)=t^2+t+16

and whose initial position is 10.

Possible Answers:

$\frac{t^3}{3}+\frac{t^2}{2}+16t+10$

${t^3}+{t^2}+16t+10$

$\frac{t^3}{3}+\frac{t^2}{2}+10$

$\frac{t^3}{3}+\frac{t^2}{2}+16t+C$

Correct answer:

$\frac{t^3}{3}+\frac{t^2}{2}+16t+10$

Explanation:

The position function describing any object is the antiderivative of the velocity function (in other words, velocity is the derivative of position).

So, we first integrate the velocity function:

$s(t)=\int v(t)dt= \int (t^2+t+16)dt=\frac{t^3}{3}+\frac{t^2}{2}+16t+C$

The following rule was used for integration:

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

Now, to determine the constant of integration, we use our initial condition given,

s(0)=10

Plugging this into our function, we get

0+0+0+C=10

C=10

Our final answer is

$\frac{t^3}{3}+\frac{t^2}{2}+16t+10$

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

The function describing the acceleration of a spacecraft with respect to time is

a(t)=4t

Determine the function describing the position of a spacecraft given that the initial acceleration is 0, the initial velocity is 3, and the initial position is 9.

Possible Answers:

$\frac{2t^3}{3}+3t$

2t^2+3

$\frac{2t^3}{3}+3t+9$

$\frac{2t^3}{3}+9$

Correct answer:

$\frac{2t^3}{3}+3t+9$

Explanation:

To find the position function from the acceleration function, we integrate the acceleration function to find the velocity function, and integrate again to get the position function:

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

The integral was found using the following rule:

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

To find the constant of integration, we use the initial velocity condition given:

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

Now, after replacing C with the known value, we integrate the velocity function to get the position function:

$s(t)= \int v(t)dt=\int( 2t^2+3 )dt = \frac{2t^3}{3}+3t+C$

The same rule of integration was used as above.

We use the same procedure to solve for C, too, only this time using the initial position condition:

s(0)=9=0+0+C=C

Our final answer is

$s(t)=\frac{2t^3}{3}+3t+9$

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Example Question #4 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Given a particle with an acceleration at time to be a(t)=4 . With initial conditions v(0)=4 and p(0)=-4 where v(t) is the velocity at time , and p(t) is position of the particle at time .

Find the position at time .

Possible Answers:

p(t)=t^2+t+4

p(t)=2t^2-4t

p(t)=2t^2+4t-4

p(t)=2t^2

p(t)=2t^2+t-4

Correct answer:

p(t)=2t^2+4t-4

Explanation:

We first must establish the following relationship

p''(t)=a(t) and p'(t)=v(t)

We now may note that

$\int a(t)dt=v(t)+C$ or $\int4dt=4t+C=v(t)$

Since v(t)=3t+C

We must plug in our initial condition v(0)=4

Therefore our new velocity equation is v(t)=4t+4

We now may similarly integrate the velocity equation to find position.

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

Plugging in our second initial condition p(0)=-4

We find our final equation to be:

p(t)=2t^2+4t-4

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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 #17 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #17 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #21 : Applications Of Antidifferentiation

Example Question #21 : Solving Separable Differential Equations And Using Them In Modeling

Example Question #21 : Applications Of Antidifferentiation

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

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

Example Question #4 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

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