Applications of antidifferentiation - AP Calculus AB

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.

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

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.

At seconds we are told the velocity is equal to .

The average value of a function on a given interval is given by the following function:

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

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:

The following rule was used for integration:

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

Plugging this into our function, we get

Our final answer is

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:

The integral was found using the following rule:

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:

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

We first must establish the following relationship

and

We now may note that

or

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.

Plugging in our second initial condition

We find our final equation to be:

Find the integral which satisfies the specific conditions of

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.

Using these rules, we can find our answer:

Will become:

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

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

Now, to solve, simply replace c with 12.12

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

The rules of integration used were

,

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

Our final answer is

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

The following rule of integration was used:

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

Our final answer is

In this problem, letting denote the position of the particle and denote the velocity, we know that . Integrating and working backwards we have,

Plugging in our initial condition, , we see immediately that .

Repeating the process again for , we find that

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

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))

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

$y=Ce^{0.043t}$

Plug in the initial condition above to see that .

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}$$}

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}$x+\frac{1}{3}$

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

, where is a constant

, where is a constant

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

Therefore, $y=2e^x$.

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

Use the quotient rule.

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

.

Then, integrate each side of the rate law, bearing in mind that will range from to , and time will range from to :

After integrating each side, the equation becomes:

.

The left side has to be evaluated from to , and the right side is evaluated from to :

. This becomes:

.

Finally, rearranging gives:

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

Taking the derivative of gives you .

Taking the derivative of gives you .

Finally taking the derivative of gives you .

Therefore .

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

Since point is on the curve, .

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