## Master AP Calculus AB & BC

**Part II. AP CALCULUS AB & BC REVIEW**

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**CHAPTER 5. Advanced Topics in Differentiation**

**POLAR DERIVATIVES (BC TOPIC ONLY)**

One of the defining qualities of polar equations is their similarity to parametric equations. Remember that any polar function r = f(θ) can be expressed parametrically by x = rcos θ, y = rsin θ. Therefore, you differentiate polar equations using essentially the same method outlined for parametric equations:

The derivative of the y component divided by the derivative of the x component, both with respect to θ, as both should contain θ’s.

The only difference between this and the other formula is that the independent variable is θ instead of t, the typical parameter in parametric equations. While the mathematics involved is not too difficult, there are a lot of places to make mistakes; this is the only thing that makes polar differentiation tricky. Make sure to proceed slowly and cautiously.

Example 10: Find the slope of the tangent line to r = 2 + 3cos θ if θ = π/3.

Solution: First, express r parametrically as x = rcosθ and y = rsinθ:

NOTE. cos2θ = cos^{2}θ - sin^{2}θ according to trigonometric double angle formulas. Check Chapter 2 if you don t remember these.

Now, differentiate to get You’ll have to use the Product Rule to find dy/dθ:

To find the derivative at θ = π/3 substitute that into dy/dx:

No particular step of this problem is difficult, but a single incorrect sign could throw off all your calculations.

Example 11: At what values of 0, 0 ≤ θ ≤ π/2, does r = 2cos (3 θ) have vertical or horizontal tangent lines?

Solution: Questions regarding tangent lines lead you right to the derivative; begin with parametric representation: x = 2cos (3 θ) cos θ, y = 2cos (3 θ) sin θ). Now take the derivative of each with respect to θ in order to build dy/dx: each will require the Product Rule:

TIP. If dx/dθ and dy/dθ are 0 at the same time, you cannot draw any conclusions concerning horizontal and vertical asymptotes.

Remember, horizontal tangents occur when the numerator of the slope is 0 (but the denominator isn’t), and vertical tangents occur when the denominator of the slope is 0 (but the numerator isn’t). Use your calculator to solve these equations (but you’ll have to set it back into rectangular mode first). dx/dθ equals 0 at θ = 0, .912, and π/2; the derivative will not exist at these points due to vertical tangent lines, dy/dθ equals 0 at θ = .284 and 1.103; the derivative will be zero for these values, indicating horizontal tangent lines.

**EXERCISE 5**

Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.

YOU MAY USE YOUR GRAPHING CALCULATOR FOR PROBLEMS 3 AND 4.

1. If r(θ) = 1 + sin θ, where is r'(θ) defined on [0,2π]?

2. Find the equation of the tangent line to r = tan θ when θ = 11π/6.

3. If r(θ) = 3 - sin (3θ), at what values of θ is r'(θ) = 1?

4. Find the slopes of the tangent lines to r = a ∙ sin(2θ) (a > 0) at the four points furthest from the origin (as indicated in the graph below).

**ANSWERS AND EXPLANATIONS**

1. First, convert the polar equation to rectangular equations using the formulas x = r(θ)cosθ and y = r(θ)sinθ. We have r(θ) = 1 + sinθ. So,

Now, calculate the derivatives of the x- and y-components: dy/dθ = cos θ + 2sin θ cos θ and dx/dθ = -sin θ + cos 2θ. The derivative, dy/dx, will be defined wherever dx/dθ ≠ 0, so set it equal to zero to find these points:

The first two values correspond to vertical tangent lines, and the final value corresponds to a sharp point on the graph, (as shown in the graph below).

2. In parametric form, r = tan θ becomes x = tan θ cos θ = sin θ, and y = tan θ sin θ. Use these to find the rectangular coordinates of the point of tangency (when θ = 11π/6). If you plug θ = 11π/6 into both, you get the coordinate Now, you need to find the slope of the tangent line. To do so, find the derivatives of x and y (using Product Rule for y'):

Therefore

Therefore, the equation of the tangent line is as follows:

That was truly an ugly problem, but doing it without your calculator toughened you up some—admit it.

3. Once again, it is important to express the polar equation in parametric form: x = 3 cos θ — sin 3θ cos θ, y = 3 sin θ — sin 3θ sin θ. Take the derivatives of each to get below:

TIP. Even though your graphs are done in polar mode, all equation-solving with the calculator (using x-intercepts) requires that you switch back to rectangular mode first.

You want to find when that big, ugly thing equals one, so carefully type it into your calculator and solve that equation.

There are four solutions, according to the graph below, and they are θ = 1.834, 2.699, 3.142 (or π), and 5.678.

4. You have to decide what points will be the furthest from the origin; in other words, what’s the largest a ∙ sin(2 θ) can be? To start with, sin (2 θ) can be no larger than 1 and no smaller than -1 (the range of sin x). Therefore, the graph of a ∙ sin(2 θ) can be no further than a ∙ 1 = a units from the origin. (Note that a distance of-a from the origin is the same, just in the opposite direction.) Set the equation equal to ±a to find out which values of θ give these maximum distances:

So, you need to find the derivatives at these values. Note that the derivatives will not change regardless of a’s value. To convince yourself of this, you may want to draw a couple of graphs with different a values. There’s no shame in using your calculator to evaluate these derivatives since this is a calculator-active question. (If you’re not sure how to do that, make sure to read the technology section at the end of the chapter.) The derivative is 1 for θ = 3π/4 and 7π/4, and the derivative is -1 for θ = π/4 and 5π/4.