# AP Calculus AB : Techniques of antidifferentiation

## Example Questions

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### Example Question #1 : Techniques Of Antidifferentiation

A projectile is shot up from a platform  above the ground with a velocity of . Assume that the only force acting on the projectile is gravity that produces a downward acceleration of . Find the velocity as a function of .

Explanation:

with initial conditions

Separate velocity variables and solve.

Plug in intial conditions

### Example Question #2 : Techniques Of Antidifferentiation

A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of  after  seconds. What is its velocity 500 ft into the air?

Explanation:

The bullet will be at a height of 500 ft on the way up and on the way down.

We use the position equation  to solve for how long it will take to reach a height = 500. and  seconds.

We then plug that into the velocity equation, which is the derivative of the position function. .

We can see that plugging in the value of  yields  and  yields . The positive and negative values of velocity indicates the up and down direction of travel.

### Example Question #3 : Techniques Of Antidifferentiation

The position of a particle as a function of time is given below:

At what values of  does the particle change direction?

Explanation:

In order to find the point at which the particle changes direction, we must determine whenever the velocity of the particle changes sign (from positive to negative, or from negative to positive).

We will need to have the function of the particle's velocity before we can determine where the velocity changes sign. Because the velocity is the derivative of position with respect to time, we can write the function for velocity, , as follows:

If we set , then we can determine the points where it can change sign.

The possible points where  will change signs occur at . However, we need to check to make sure.

First, we can try a value less than 2, such as 1, and then a value between 2 and 4, such as 3. We will evaluate  at  and  and see if the sign of the velocity changes.

Thus,  is indeed a point where the velocity changes sign (from positive to negative). This means that the particle does in fact change direction at .

Lastly, we will evaluate the velocity at a value of  larger than 4, such as 5.

The sign of the velocity has switched back to positive, so the particle does indeed change direction at .

### Example Question #4 : Techniques Of Antidifferentiation

The speed of a car traveling on the highway is given by the following function of time:

Note that

What does this mean?

The car is not decelerating at time .

The car's speed is constantly changing at time .

The car is not accelerating at time .

The car takes  seconds to reach its maximum speed.

The car is not moving at time .

The car is not moving at time .

Explanation:

The function  gives you the car's speed at time . Therefore, the fact that  means that the car's speed is  at time . This is equivalent to saying that the car is not moving at time . We have to take the derivative of  to make claims about the acceleration.

### Example Question #5 : Techniques Of Antidifferentiation

A jogger leaves City  at.  His subsequent position, in feet, is given by the function:

,

where  is the time in minutes.

Find the velocity of the jogger at 15 minutes.

Explanation:

To find velocity, one has to use the first derivative of :

.

Note the units have to be ft/min.

### Example Question #1 : Techniques Of Antidifferentiation

Consider the function

Find the minimum of the function on the interval .

Explanation:

To find potential minima of the function, take the first derivative of  using the power rule.

Set the derivative to 0:

We solve for  to obtain  and then plug in 0.5 into the original function to obtain the answer of

We can double check that  is indeed a minimum by using the second derivative test

which means the function is concave up, so that the point we found is a minimum.

### Example Question #6 : Techniques Of Antidifferentiation

Find the x coordinate of the minimum of  on the interval

Explanation:

First, find the derivative, which is:

Set that equal to 0 to find your critical points:

Evaluate f at each critical point and endpoint over [-2, 2]:

The least of those numbers is the minimum. Because f(-2)=-32, that is the minimum.

### Example Question #1 : Calculus I — Derivatives

What is the local minimum of  when ?

The y-value is constant throughout that range.

There is no local minimum in that range.

Explanation:

To find the maximum, we need to look at the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat  as  since anything to the zero power is one.

Notice that  since anything times zero is zero.

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Because we want the MINIMUM, we want to see where the derivative changes from negative to positive.

Notice that  has a root when . In fact, it changes from negative to positive at that particular point. This is the local minimum in the interval .

### Example Question #7 : Techniques Of Antidifferentiation

Integrate,

Explanation:

Integrate

1) Apply the sum rule for integration,

2) Integrate each individual term and include a constant of integration,

Further Discussion

Since indefinite integration is essentially a reverse process of differentiation, check your result by computing its' derivative.

This is the same function we integrated, which confirms our result. Also, because the derivative of a constant is always zero, we must include "C" in our result since any constant added to any function will produce the same derivative.