Applications of antidifferentiation - AP Calculus AB

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Question

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?

Answer

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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Question

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:

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.)

Answer

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 as the constant of integration since the function is a velocity and at initially, at , the velocity is equals the constant of integration.

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

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

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Question

Find the average value of the function on the interval $[0, \pi].$

Answer

The average value of a function on a given interval 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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Question

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

and whose initial position is 10.

Answer

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,

Plugging this into our function, we get

Our final answer is

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

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Question

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

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.

Answer

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:

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:

Our final answer is

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

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Question

Given a particle with an acceleration at time to be . With initial conditions and where is the velocity at time , and is position of the particle at time .

Find the position at time .

Answer

We first must establish the following relationship

and

We now may note that

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

Since

We must plug in our initial condition

Therefore our new velocity equation is

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

We find our final equation to be:

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Question

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

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

Answer

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:

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

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

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Question

Find the velocity function given the following information:

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

Answer

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:

Our final answer is

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

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Question

What is the position function if the initial position is 0 and the velocity function is given by ?

Answer

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:

Our final answer is

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Question

A particle at the origin has an initial velocity of . If its acceleration is given by , find the position of the particle after 1 second.

Answer

In this problem, letting denote the position of the particle and denote the velocity, we know that . 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, , we see immediately that .

Repeating the process again for , 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, (we started at the origin) we see that . This gives us a final equation

. The problem asks for which is simply

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Question

Find (dy/dx).

sin(xy) = x + cos(y)

Answer

The first step of the problem is to differentiate with respect to (dy/dx):

cos(xy)\[(x)(dy/dx) + y(1)\] = 1 – sin(y)(dy/dx)

*Note: When differentiating cos(xy) remember to use the product rule. (xy' + x'y)

Step 2: Clean the differentiated problem up

cos(xy)(x)(dy/dx) + cos(xy)y = 1 – sin(y)(dy/dx)

cos(xy)(x)(dy/dx) + sin(y)(dy/dx) = 1 – cos(xy)y

Step 3: Solve for (dy/dx)

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

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Question

Suppose $1000 is invested in an account that pays 4.3% interest compounded continuously. Find an expression for the amount in the account after time .

Answer

The differential equation is $\frac{dy}{dt}=0.043y$ , with boundary condition $y(0)=1000$ .

This is a separable first order differential equation.

$\frac{1}{y}dy=0.043dt$

Integrate both sides.

$ln(y)=0.043t+c$

$e^{ln(y)}=e^{0.043t+c}$

$y=Ce^{0.043t}$

Plug in the initial condition above to see that .

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Question

Find $y^{'}$ if

$y=\frac{ln(x)}{x^{3}}$

Answer

The answer is

$y'=\frac{1-3ln(x)}{x^{4}}$

$y=\frac{ln(x)}{x^{3}}$

$y'=\frac{(\frac{1}{x})x^{3}-ln(x)(3x^{3})}{x^{6}}$

$y'=\frac{x^{2}(1-3ln(x))}{x^{6}}$

$y'=\frac{1-3ln(x)}{x^{4}}$

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Question

Find the equation of the normal line at on the graph $y=x^{3}-6x+4$ .

Answer

The answer is $y=\frac{-1}{6}x+\frac{1}{3}$ .

$y=x^{3}-6x+4$

$y'=3x^{2}-6$

Now plug in .

$y'=3(2)^{2}-6 = 6$ now we know 6 is the slope for the tangent line. However, we aren't looking for the slope of the tangent line. The slope of the normal line is the negative reciprocal of the tangent's slope; meaning the slope of the normal is $\frac{-1}{6}$ . Now find the equation of the normal line.

$y-0=\frac{-1}{6}(x-2)$

$y=\frac{-1}{6}x+\frac{1}{3}$

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Question

Find the solution to the equation $y'=y$ at $x=2$ with initial condition $y(0)=2$ .

Answer

First, we need to solve the differential equation of $y'=y$ .

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

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

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

, where is a constant

, where is a constant

To find , use the initial condition, , and solve:

Therefore, $y=2e^x$ .

Finally, at , $y(2)=2e^2$ .

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Question

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

What is the derivative of ?

Answer

Use the quotient rule.

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

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Question

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 .

Answer

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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Question

Differentiate .

Answer

While differentiating, multiply the exponent with the coefficient then subtract the exponent by one.

$y'=(3\cdot3)x^{3-1}+(5\cdot1)x^{1-1}+(7\cdot0)$

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Question

If what is ?

Answer

Taking the derivative of gives you .

Taking the derivative of gives you .

Finally taking the derivative of gives you .

Therefore .

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Question

Solve the differential equation: $\frac{dy}{dx}=-2xy$

Note that is on the curve.

Answer

In order to solve differential equations, you must separate the variables first.

$\frac{dy}{y}=-2xdx$

$\int \frac{dy}{y}= \int -2xdx$

Since point is on the curve, .

To get rid of the log, raise every term to the power of e:

$e^{ln(y)}=e^{-x^2}e^{ln(2)}e^1$

$y=2e^{1-x^2}$

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