Mr. O'Brien's 2010-11 Period 2 AP Calculus: Scribe Post Tuesday October 19, 2010

The overall focus of this class was Implicit Differentiation, which is also known as the Chain Rule.

As usual, we first read over the blog post from the previous class. During this time, and on the subject of the derivatives of even and odd functions, Mr. O'Brien noted that sin(x) and cos(x) are also odd and even functions respectively. [which is logical, as the graph of sin(x) is clearly symmetric about the origin, a characteristic of odd functions (see below)]

Next we corrected the quiz from last class. There was brief discussion about scaling, and although Mr. O'Brien emphasized that the questions were relatively standard, get-a-3-on-your-AP-test problems, there will be a small scale, as we are in the process of learning.

Additionally, it was noted that the h-->0 and x-->c forms of finding derivatives will once again be on the quiz Thursday. As a class we're getting better with them, but they'll still be there just for reinforcement.

Assignment Questions followed quiz discussion. We reviewed pg.165 (31,53,59). At some point during the course of this review, OB reminded us that it's important to correctly annotate equations for the derivatives of functions. For example, if you're trying to find the derivative of, say,

, it's important to start off your work with either [ y'= ] or [ dy/dx= ].

Also on the topic of homework questions, Mr. O'Brien pointed out an identity we should be on the lookout for if we want a 5 on the AP Test:

$2(sin\theta)(cos\theta) = sin(2\theta)$

In the context of the problem we were solving at the time, we had an answer of:

This could easily be disguised in a multiple choice problem using the above identity. Options you might see are as follows:

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Now the real substance begins.

We started the official part of the lesson with a practice problem.

Objective: find the equation of the line tangent to "a circle centered at the origin with a radius of 4" at x = 2.

Using the equation for a circle [ (x - h)² + (y - k)² = r²], we were able to deduce that

. Let's plug this in to Geogebra.

Since we want to find a tangent line at x=2, let's plug 2 into our equation as x.

$y = \pm \sqrt12 = \pm 2\sqrt3$

The plus or minus indicates that there are in fact two tangent lines to this circle at x = 2, one at point $(2,2\sqrt 3)$ , and another at point $(2,-2\sqrt 3)$ .

Mr. O'Brien revealed that there are two methods of finding these tangent lines (one using geometry, the other calculus), the fact that there are two lines conveniently allows us to try using both.

Let's first use the geometric method to find the equation of the tangent to circle at point B, or $(2,2\sqrt 3)$ .

The rise over the run of this particular point (much like the unit circle) yields $\frac{2\sqrt 3}{2} = \sqrt 3$ for the slope of the radius here. This is a critical value, as the tangent to a circle at a given point is always perpendicular to the radius at the same point.

Since the slope of a line perpendicular to another is the inverse of the original slope negated, and since the slope of the radius at this point is $\sqrt3$ , then we can conclude that the slope of the tangent here will be $\frac{-1}{\sqrt 3}$ .

Since we know that one of the points on this tangent line is $(2,2\sqrt 3)$ , we can use these two values in conjunction with the point-slope form of a line to come up with an equation for the tangent.

[point slope form is

]

Thus, the equation for the tangent line to this circle at point $(2,2\sqrt 3)$ is:

$y-2\sqrt 3 = \frac{-1}{\sqrt3}(x-2)$

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Okay. So now we've got one of the two tangent lines to the circle at x = 2.

We'll find the other with the calculus method.

Point: $(2,-2\sqrt 3)$

Slope: ------

let's think back to the first thing we learned about derivatives -- the derivative of a function is the slope of its tangent line. Well, how convenient . . . that's exactly what we're looking for.

Okay, so we're working with a circle of equation

sweet, now let's just find the derivative, plug in 2 for x, choose the result with the positive slope and be done. easy, no?

Indeed, it would be with this FRESHMAN BINOMIAL THEOREM.

Sadly, it's not that simple. And seriously, don't do this. It's totally wrong.

$y = \pm \sqrt{(16-x^2)}$

[Since, we're looking for the tangent that touches the bottom half of the circle, we'll use $- \sqrt{(16-x^2)}$ ]

$y = - (16-x^2)^{\frac{1}{2}}$

Okay. Now we have something we can take the derivative of. We will have to use the chain rule though, which if you recall is

So,

$\frac{dy}{dx} = - \frac{1}{2} (16-x^2)^{\frac{-1}{2}} (-2x) = \frac{x}{\sqrt{16-x^2}}$

This, ladies and gentlemen, is the general form of the slope of the tangent line.

Now, let's sub in a 2 for x.

$m_{tan} = \frac{2}{\sqrt{16-4}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$

Once again we now have a a point, $(2,-2\sqrt 3)$ , and a slope, $\frac{1}{\sqrt{3}}$ .

Pop this into point-slope form, and we have the equation for the second tangent:

$y + 2\sqrt3 = \frac{1}{\sqrt3}(x-2)$

Shall we verify our two tangent equations? Let's plug them into Geogebra.

Yep. Problem solved.

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After that warm-up problem, let's learn how to use Implicit Differentiation to take the derivative of an equation with respect to x.

[as a side note, "with respect to x" means that the variable on the horizontal axis is x, or the independent variable.]

Let's use our

equation just so we can humiliate the other two methods as much as possible.

$\frac{d}{dx} [x^2 + y^2] = \frac{d}{dx} (16)$

$2x + 2y(\frac{dy}{dx}) = 0$ [note that $\frac{dx}{dx}$ is actually 1, and we don't need to include it in this notation if we don't want to]

$\frac{dy}{dx} = \frac{-2x}{2y}$

$\frac{dy}{dx} = \frac{-x}{y}$

Could it really be so simple? Did we really just solve that entire problem in four steps? Well, let's compare the slope that this derivative equation gives us to the slope that we know the tangent to the circle should have at $(2,-2\sqrt 3)$ .

$m = \frac{-x}{y} = \frac{-2}{-2\sqrt3} = \frac{1}{\sqrt3}$

Yep -- same slope. And again, if we had started with this method and wanted to finish the problem, we would use this slope and the point we know to be on the tangent, $(2,-2\sqrt 3)$ , to quickly and efficiently produce the tangent's official equation.

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Okay, one last example problem.

Let's find the tangent line to

at the point (-1,2)