# AP Calculus BC : Derivatives of Parametric, Polar, and Vector Functions

## Example Questions

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### Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Find the derivative of the following set of parametric equations:

Explanation:

We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

### Example Question #9 : Derivatives Of Parametrics

Solve for  if  and .

None of the above

Explanation:

We can determine that   since the  terms will cancel out in the division process.

Since  and , we can use the Power Rule

for all  to derive

and  .

Thus:

.

### Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Find the derivative of the following polar equation:

Explanation:

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to . This gives us:

Now that we know dr/d, we can plug this value into the equation for the derivative of an expression in polar form:

Simplifying the equation, we get our final answer for the derivative of r:

### Example Question #1 : Derivatives Of Polar Form

Find the derivative  of the polar function .

Explanation:

The derivative of a polar function is found using the formula

The only unknown piece is . Recall that the derivative of a constant is zero, and that

, so

Substiting  this into the derivative formula, we find

### Example Question #1 : Derivatives Of Polar Form

Find the first derivative of the polar function

.

Explanation:

In general, the dervative of a function in polar coordinates can be written as

.

Therefore, we need to find , and then substitute  into the derivative formula.

To find , the chain rule,

, is necessary.

We also need to know that

.

Therefore,

.

Substituting  into the derivative formula yields

### Example Question #11 : Derivatives Of Polar Form

What is the derivative of ?

Explanation:

In order to find the derivative   of a polar equation , we must first find the derivative of with respect to  as follows:

We can then swap the given values of  and  into the equation of the derivative of an expression into polar form:

Using the trigonometric identity , we can deduce that . Swapping this into the denominator, we get:

### Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Given that  . We define its gradient as :

Let  be given by:

What is the gradient of ?

Explanation:

By definition, to find the gradient vector , we will have to find the gradient components. We know that the gradient components are that partial derivatives.

We know that in our case we have :

To see this, fix all other variables and assume that you have only  as the only variable.

Now we apply the given defintion , i.e,

with :

this gives us the solution .

### Example Question #3 : Derivatives Of Parametric, Polar, And Vector Functions

Let .

We define the gradient of as:

Let .

Explanation:

We note first that :

Using the Chain Rule where  is the only variable here.

Using the Chain Rule where  is the only variable here.

Continuing in this fashion we have:

Again using the Chain Rule and assuming that  is the variable and all the others are constant.

Now applying the given definition of the gradient we have the required result.

### Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Let

What is the derivative of ?

Explanation:

To find the derivative of this vector, all we need to do is to differentiate each component with respect to t.

Use the Power Rule and the Chain Rule when differentiating.

is the derivative of the first component.

of the second component.

is the derivative of the last component . we obtain then:

### Example Question #151 : Vector

Explanation:

In general:

If ,

then

Derivative rules that will be needed here:

• Taking a derivative on a term, or using the power rule, can be done by doing the following:
• When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term:
• Special rule when differentiating an exponential function:  , where k is a constant.

In this problem,

Put it all together to get

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