Applications of Derivatives - AP Calculus AB

To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,

Plugging in , we get

Use the power rule on each term of the polynomial to get the derivative,

Now we plug in

We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:

In order to find the derivative of , we will need to employ the Chain Rule.

We can factor out a 2x to make this a little nicer to look at.

Now we must evaluate the derivative when x = .

The answer is .

To find the derivative of the function, we use implicit differentiation, where we always treat y as a function of x, and denoting any derivative of y with respect to x as

The rate of change of y is also the derivative of y.

Differentiate the function given.

You should get

The acceleration function of a particle with respect to time is the derivative of the velocity function of the particle.

Find this derivative by first using the product rule:

Apply the product rule:

Evaluate :

feet per second squared.

To find the rate of change of the diameter, we must relate the diameter to something we do know the rate of change of: the surface area.

The surface area of the top side of the pizza dough is given by

The rate of change, then, is found by taking the derivative of the function with respect to time:

Solving for the rate of change of the radius at the given radius, we get

inches/sec

Now, we relate the diameter to the radius of the pizza dough:

Taking the derivative of both sides with respect to time, we get

Plugging in the known rate of change of the radius at the given radius, we get

inches/sec

We could have found this directly by writing our surface area formula in terms of diameter, however the process we used is more applicable to problems in which the related rate of change is of something not as easy to manipulate.

To determine the rate of change of the circumference at a given radius, we must relate the circumference rate of change to the rate of change we know - that of the volume.

Starting with the equation for the volume of the spherical balloon,

we take the derivative of the function with respect to time, giving us the rate of change of the volume:

The derivative was found using the following rules:

,

The chain rule was used when taking the derivative of the radius with respect to time, because we know that it is a function of time.

Solving for using our known at the given radius, we get

Now, we use this rate of change and apply it to the rate of change of the circumference, which we get by taking the derivative of the circumference with respect to time:

Solving for the rate of change of the circumference by plugging in the known rate of change of the radius, we get

To determine the rate of the change of the angle opposite to the base of the given right triangle, we must relate it to the rate of change of the base of the triangle when the triangle is a certain area.

First we must determine the length of the base of the right triangle at the given area:

Now, we must find something that relates the angle opposite of the base to the length of the base and height - the tangent of the angle:

To find the rate of change of the angle, we take the derivative of both sides with respect to time, keeping in mind that the base of the triangle is dependent on time, while the height is constant:

We know the rate of change of the base, and we can find the angle from the sides of the triangle:

Plugging this and the other known information in and solving for the rate of change of the angle adjacent to the base, we get

radians per minute

To answer this question, one first needs to find and then find the critical points of the function (i.e. where . Finally, one would need to determine the sign of for the intervals between the critical points.

For the given function:

.

Therefore, when and . So, the intervals to consider are:

To determine the sign of , pick any number for the given interval and evaluate at that number.

Therefore, is increasing on the intervals and since is greater than zero on these intervals.

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Plug in our given value.

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use as our expression.

For this problem, we're going to say that since, as stated before, anything to the zero power is one.

Notice that as anything times zero is zero.

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

To find the intervals where the function is increasing, we need to find the points at which its slope changes from positive to negative and vice versa. The first derivative, which is the slope at any point, will help us.

First, we find the derivative of , using the power rule for each term. Recall that the power rule says

Also, the constant multiple rule will apply to the coefficients of each term. The constant multiple rule simply says that any constant factor of a term will "carry" to the derivative of that term. For example:

Lastly, the derivative of a constant is zero. This will result in the last term, , dropping off as we take the derivative.

Applying these rules, we find the derivative

Where the derivative is positive (blue line in graph), the tangent line to the original function is angled up. Where the derivative is negative (red line in graph), the slope of the tangent line is angled down.

The points where the tangent line's slope transitions from negative to positive or vice versa, are called the critical points. At these points, the tangent line becomes a horizontal line with a slope of zero (green line in graph). In other words, the "critical points" occur when the derivative is zero. These points will be the endpoints of our intervals of increasing and decreasing. To find the critical points, we will set the derivative equal to zero and solve for x. In this problem factoring is the best method:

Now that we have found the critical points, we need to know whether the original function is increasing or decreasing in the intervals between them. We will do so by testing a point in each interval and determining whether the derivative is positive or negative at that point. This is called the first derivative line test.

For the interval , we will test .

(Note: I will use the factored form of the derivative, but we could also use the polynomial version. Both will give the same result)

Since the derivative is negative at this point, we can conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

For the interval , we will test .

Since the derivative is positive at this point, we can again conclude that the derivative is positive for the whole interval. Thus, the original function is increasing on .

For the interval , we will test .

Since the derivative is negative at this point, we can again conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

Lastly, for the interval , we will test .

Since the derivative is positive at this point, we know that the derivative is positive for the whole interval. Thus the original function is increasing on .

From these 4 results, we now know the answer. The function is increasing on the intervals .

To find the coordinates of the local extrema of a function, we need to find the critical points of its first derivative.

Since is a polynomial, we can find its derivative term by term. The first 3 terms can be differentiated using the power rule, , and the constant multiple rule, .

The last term is a constant, and its derivative is zero.

Applying these rules, we find the first derivative:

Now we need to find the critical points. To do this, we set the first derivative equal to zero and solve. Factoring is the best method in this problem.

Now that we have the critical points, we need to determine for each one, whether it is a maximum, minimum, or neither. We use the first derivative line test to determine this.

For , we will test the interval before it, , and the interval after it, , and find whether they are increasing or decreasing.

For the interval , we will test and find whether is positive or negative.

Since is positive, the original function is increasing before the critical point, .

Now we will test the interval after .

For the interval, , we will test and find whether is positive or negative.

Since is negative, the original function is decreasing in the interval following .

Since the function is increasing before , and decreasing afterward, we can conclude that a maximum occurs at .

Now we find the value of this maximum, .

Thus is a local maximum.

Now we will determine whether a maximum or minimum occurs at .

We know that is decreasing before , but we still need to determine what happens afterward.

For the interval , we will test , and find whether is positive or negative.

Since is positive, the original function is increasing following .

Since is decreasing before, and increasing after , we can conclude that a minimum occurs at .

Now we need to find the value of this minimum, .

Thus is a local minimum.

So our answer is:

is a local maximum.

is a local minimum.

Concavity refers to the "curving" of the function. While the first derivative describes when the function is increasing or decreasing (instantaneous rate of change of ), the second derivative describes concavity, the instantaneous rate of change of .

The first derivative is like velocity, (moving forward or backward), while the second derivative is like acceleration (speeding up or slowing down).

Since we need to find the intervals of concavity, we will find the second derivative and work with it.

First we must find the first derivative using the power rule and constant multiple rule for each term of .

This gives:

Now to find the second derivative, we take the derivative of .

The same derivative rules apply:

Now that we have the second derivative, , we must find its critical points. We do this by setting , and solving.

The best method for this case is factoring.

The greatest common factor is ,

Inside the parentheses is a quadratic expression that can be factored like so:

Setting each factor equal to zero we find the following:

Now we know the critical points for the second derivative. To find the intervals of concavity, we test a point in each interval around these critical points and find whether the second derivative is positive or negative in that interval.

For the interval , we can test .

Using the factored form of the second derivative is easier than using the polynomial form, since the arithmetic involves fewer large numbers.

Since is negative, the original function is Concave Down in the interval .

Now for the interval , we can test .

We will combine the fractions inside the parentheses by getting the common denominator.

Multiplying the fractions gives:

Since is positive, the original function is Concave Up on the interval .

For the interval , we can test .

Since is negative, the original function is Concave Down on the interval .

Finally, for the interval , we can test .

Since is positive, the original function is Concave Up on the interval .

Summarizing the results, the intervals of concavity are:

Concave Down:

Concave Up:

On the domain , we know that the derivative begins positive, and because the concavity is positive, we know the derivative is increasing. Thus, the derivative stays positive for this entire interval, and the function increases from 2 to 4. Thus, must be greater than .

In the case of the interval , we know that the derivative is increasing, but it starts out negative. Thus, perhaps the derivative only increased from -1 to -0.5 in this interval, and the function would have decreased the entire time. In this case, would be less than , so we can't really say anything about these values.

For the remaining two, there's not any clear way to relate the functions at and . While we know needs to be bigger than 1, we don't know by how much. Similarly, while we know needs to be bigger than -1, we don't know by how much. Thus, it's completely possible that and .

As for and , we know even less. In between the two intervals, our function could have shot up a million, or shot down by the same amount. Thus, there's no safe comparison we can make between these two values.

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at . In order to find the tangent line, we will need to evaluate the derivative of at .

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

$f'(x)=2x-(-1)(1-x^2$$)^{-2}$(-2x)

$f'(x)=2x-2x(1-x^2$$)^{-2}$

$f'(2)=2(2)-2(2)(1-2^2$$)^{-2}$

f'(2)=4-4($\frac{1}{9}$)=\frac{32}{9}$

The slope of the tangent line at is . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at , it will pass through the point . Be careful to use the original equation for , not its derivative.

$f(2)=2^2$$-(1-4)^{-1}$=4-(-$\frac{1}{3}$)=\frac{13}{3}$

The normal line has a slope of and passes through the piont . We can now use point-slope form to find the equation of the normal line.

y-$\frac{13}{3}$=-$\frac{9}{32}$(x-2)

Multiply both sides by .

96y-416=-27(x-2)

27x + 96y = 470

The answer is 27x + 96y = 470.

We simply differentiate both sides of the equation

(Don't forget the chain rule)

Now we solve for

Implicit differentiation is very similar to normal differentiation, but every time we take a derivative with respect t , we need to multiply the result by We also differentiate the entire equation from left to right, including any numbers. Then, we solve for that for our final answer.

This is a related rates problem.

Since the problem gives the time for one orbit, we can find the angular speed of the point. The angular speed is simply how many radians the particle travels in one second. We find this by dividing the amount of radians in one revolution, , by the time it takes to travel one revolution, 8 seconds.

This gives us the change in the angle with respect to time, .

Now we need to relate the position to the angle, . Recall that

, where r is the radius.

Since the radius is given as 1 unit, we can write this equation as

.

Now we take the derivative of both sides with respect to time, using implicit differentiation. Remember that we use the chain rule for any variable that is not . This gives,

.

Now we have a formula that relates the horizontal speed of the particle at a instant in time, , to the angle above the positive x axis and angular speed at that same instant. We are told to find how fast the x coordinate is changing when the angle, is radians above the positive x axis. So we will plug in for . However we also need to know . Fortunately, we already found it. It is the angular speed, radians/second.

Plugging this information in, we get

This is the answer. The negative makes sense, because the point is traveling counter-clockwise. Hence, it is moving left when the angle is radians.

Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to , we need to implicitly write after it. Hence, the name of this method. Then, we solve for