# AP Calculus AB : Applications of antidifferentiation

## Example Questions

### Example Question #18 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

and at

Explanation:

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

Next, we integrate both sides:

The integrals were solved using the following rules:

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

Now, we exponentiate both sides to solve for y:

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

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

Using the given condition, we can solve for C:

### Example Question #21 : Applications Of Antidifferentiation

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

,

where  is the concentration of compound  at a given time, .  Which one of the following equations describes  as a function of time?  Let  be the concentration of compound  when .

Explanation:

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:

### Example Question #11 : Solving Separable Differential Equations And Using Them In Modeling

Given that  and , solve for . What is the value of ?

1

1

Explanation:

This is a separable differential equation. The simplest way to approach this is to turn  into , and then by abusing the notation, "multiplying by dx" on both sides.

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

Integrating both sides, we find

Here, the first integral is found by using substitution of variables, setting . 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.

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

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

As  is just another arbitrary constant, we can relabel this as C, or  if you prefer. Solving for y gets us

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

Leaving us with a final equation of

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

### Example Question #22 : Applications Of Antidifferentiation

Solve the separable differential equation:

given the condition at

Explanation:

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

Next, we integrate both sides:

The integrals were solved using the identical rules.

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

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

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

### Example Question #23 : 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?

1.5s

1s

5s

3s

0s

1s

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) = –15t+ 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

### 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:

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.)

Explanation:

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

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

Find the average value of the function  on the interval

Explanation:

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:

### 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

and whose initial position is 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:

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

### Example Question #4 : 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

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.

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:

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:

### Example Question #5 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

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 .

Explanation:

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: