with initial conditions

Separate velocity variables and solve.

Plug in intial conditions

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

We use the position equationto solve for how long it will take to reach a height = 500.andseconds.

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

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

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 wherewill 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 evaluateatandand 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 oflarger than 4, such as 5.

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

The answer isand.

The functiongives you the car's speed at time. Therefore, the fact thatmeans that the car's speed isat time. This is equivalent to saying that the car is not moving at time. We have to take the derivative ofto make claims about the acceleration.

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

.

Note the units have to be ft/min.

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

Set the derivative to 0:

We solve forto obtainand then plug in 0.5 into the original function to obtain the answer of

We can double check thatis 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.

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.

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 treatassince anything to the zero power is one.

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

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.