Find the slope of the tangent line of the polar curve r = 2 cos θ r=2\cos\theta at θ = π 3 \theta=\frac{\pi}{3} .

So here we have an interesting problem where we're asked to find the slope of the tangent line to the polar curve at r equals two cosine theta at a theta value of pi over three. How do we go about solving this problem? Well, notice we're asked to find the slope of the tangent line. The slope of the tangent line is simply a derivative at a certain point, and so that you can see we're dealing with a polar curve, we're going to need to be taking the derivative of a polar curve. So how do we go about doing that? Well, what we're looking for when we're finding derivatives of polar curves is dy over dx, and since we're in terms of r and theta, well, what it's gonna look like is this. It's gonna be the derivative of y with respect to theta over the derivative of x with respect to theta. Now we know how to convert from polar to Cartesian coordinates. Recall that x is just going to be f of theta times the cosine of theta, and then y is going to be f of theta times the sine of theta. So what we can actually do is write out a more detailed formula right here. So we have dy over d theta, which is gonna be the derivative right here, and that's going to be using the product rule f prime of theta times the sine of theta plus f of theta cosine theta. And we've derived this before, so I'm just gonna write this all out here, and it's going to be over the derivative of x with respect to theta, which is going to be f prime of theta cos theta plus or in this case, that's actually gonna be minus, and we're going to have f of beta sine theta. So this is what it looks like when we take the derivatives of y and x with respect to theta and then divide them, which will give us this formula right about here. So now that we have this formula derived, let's see if we can directly plug things into this formula to get a result for dy over dx. Well, notice that we need to know what f of theta is, and f of theta is going to be this piece right here. It's going to be this r. So this is going to be our f of theta. And if I want to find f prime of theta, that's going to be the derivative of two cosine theta. Well, I know that this two is going to stay the same, and the derivative of cosine is negative sine. So we get negative two sine of theta for our derivative. Now at this point, since I know what my x and y are, and we had went ahead and found our f prime of theta, and we have f of theta, let's just plug everything into this formula. So we're going to start with f prime of theta, which I can see right here is negative two sine theta. That's going to be multiplied by the sine of theta. We're going to have plus f of theta, which I can see right here is two cos theta multiplied by the cosine of theta. Now all of this is going to be divided by, well, what we have on bottom right here. So we have f prime of theta, which I can see here is negative two sine theta, then we multiply that by the cosine. And And then we're going to have minus, and then we have f of theta, which is just gonna be this two cos theta up here times the sine of theta. So that's what our dy over dx is equal to. Now at this point, things look really confusing, so let's see if we can get this to look any more simple. Well, I know that we can combine these sines to get sine squared, so we get negative two sine squared theta. Now I also know we can combine these cosines here. That's gonna give us two cosine squared theta. Now all of this is going to be divided by, and then in this case, we're going to have negative two sine theta cos theta. And two sine theta cos theta, that's actually a trigonometric identity. That's a double angle identity. That's gonna be the sine of two theta. I just have to keep this negative on the outside here. And then we're going to have minus, and then notice we have the same double angle identity right here, two sine theta cos theta. So in this case, it's going to be once again the sine of two theta. So what we end up getting here is this case. So notice that we have this negative on bottom here, and we can actually factor that out. And, well, in this case, since we have sine two theta minus sine two theta, this is gonna be negative two sine two theta. So on top, we're gonna get negative two sine squared theta plus two cos squared theta. And then on bottom, we're going to get negative two sine two theta. Now what can I do at this point? Well, notice on top here that I have this two in common with each of these terms. And what I'm actually going to do is flip this so we have the sine squared over on this side and the cosine squared over on the other side. So I'm gonna do this all in one step. So So what's going to give me is negative, and then well, actually, no. We're gonna have the two factored out. And then in parentheses here, I'm going to have cosine squared theta minus sine squared theta. So all I really did here was I took this negative sign, I attached it to the sine squared, and then I moved the sine squared to the other side, and then I pulled this two out. And there's a reason that I did all of this right here. And then we have on bottom, negative two sine two theta. Now here's the reason I did this step. I get sine cosine squared beta minus sine squared beta right here when I do this, and that's actually another identity that we have. Whenever you see cosine squared beta minus sine squared beta, that's equivalent to cosine two theta. So So what this is actually going to come out to is two cosine two theta on top, and then it's going to be divided by negative two sine two theta on bottom. And this is interesting because clearly the two's here are going to cancel, but we'll still have a negative result that comes out of this. But cosine over sine, that's actually just cotangent. That's another identity. So all of this just comes up to negative cotangent of two theta. So this is what we get. Now we're not quite done with the problem yet, even though we did find our dy over dx, because we're trying to find the slope of the tangent line. And the slope of the tangent line is a derivative at a certain point. So we went ahead and found our derivative dy over dx, but I need to find this derivative at an angle of pi over three. That's what we're looking for. So what that means is that I'm looking for dy over dx at an angle of pi over three, that means I need to go ahead and take pi over three and plug it in for theta here. So that's gonna give us negative cotangent of two pi over three, plugging pi over three in for that angle right about there. And at this point, what you can do is look at this on a unit circle, and you should get that the cotangent of two pi over three comes out to negative one over the square root of three. Now notice here the negatives will cancel. The negative here will cancel with that one, giving us one over the square root of three. And then what I'm gonna go ahead and do is rationalize the denominator here by multiplying the top and the bottom by the square root of three. Then I get the square roots to cancel right there, giving us square root three over three, and that right there is the solution to the problem and the slope of the tangent line. So I hope you found this video helpful, and let's move on.

16. Parametric Equations & Polar Coordinates

Calculus in Polar Coordinates

Calculus

16. Parametric Equations & Polar Coordinates

Calculus in Polar Coordinates

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Multiple Choice

Find the slope of the tangent line of the polar curve $r equals 2 cosine theta$ at $θ = \frac{π}{3}$ .

$the square root of 3$

$- \frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3}$

$- \sqrt{3}$

Verified step by step guidance

Step 1: Recall that the slope of the tangent line to a polar curve is given by the formula $ \frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta} $, where $ r' $ is the derivative of $ r $ with respect to $ \theta $.

Step 2: Start by differentiating $ r = 2 \cos \theta $ with respect to $ \theta $. Using the chain rule, $ r' = \frac{d}{d\theta}(2 \cos \theta) = -2 \sin \theta $.

Step 3: Substitute $ r = 2 \cos \theta $, $ r' = -2 \sin \theta $, and $ \theta = \frac{\pi}{3} $ into the formula for $ \frac{dy}{dx} $. Evaluate $ \sin \theta $ and $ \cos \theta $ at $ \theta = \frac{\pi}{3} $: $ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} $ and $ \cos \frac{\pi}{3} = \frac{1}{2} $.

Step 4: Plug these values into the numerator $ r' \sin \theta + r \cos \theta $ and denominator $ r' \cos \theta - r \sin \theta $ of the slope formula. Simplify each term step by step.

Step 5: After simplifying the numerator and denominator, divide them to find $ \frac{dy}{dx} $, which represents the slope of the tangent line at $ \theta = \frac{\pi}{3} $.