Video: Finding the Slope of a Tangent Line to a Polar Curve

Find the slope of a tangent line to π‘Ÿ = 6 + 3 cos πœƒ at (3, πœ‹).

04:14

Video Transcript

Find the slope of a tangent line to the polar equation π‘Ÿ is equal to six plus three multiplied by the cos of πœƒ at the point three, πœ‹.

The question is asking us to find the slope. And we recall that the slope of a function is given by the rate of change of 𝑦 with respect to π‘₯. And we want to find the slope of the tangent line at the point three, πœ‹. So this is when π‘Ÿ is equal to three, and πœƒ is equal to πœ‹. Now, we recall our polar coordinate equations which tell us that 𝑦 is equal to π‘Ÿ multiplied by the sin of πœƒ. And π‘₯ is equal to π‘Ÿ multiplied by the cos of πœƒ. In fact, from the question, we’re told that π‘Ÿ is equal to six plus three multiplied by the cos of πœƒ. We can substitute this into our polar equations to get that 𝑦 is equal to six plus three multiplied by the cos of πœƒ all multiplied by the sin of πœƒ. And π‘₯ is equal to six plus three cos of πœƒ all multiplied by the cos of πœƒ.

We now recall the chain rule which tells us the derivative of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to πœƒ multiplied by the derivative of πœƒ with respect to π‘₯. So we could use this to find our slope function, if we could find the derivative of 𝑦 with respect to πœƒ and the derivative of πœƒ with respect to π‘₯. We’ve already found 𝑦 written as a function of πœƒ. So we could find the derivative of 𝑦 with respect to πœƒ. However, we don’t have πœƒ as a function of π‘₯. Instead, we have π‘₯ is a function of πœƒ. But this is okay since we know we can modify our chain rule. Instead of multiplying by the derivative of πœƒ with respect π‘₯, we can divide by the derivative of π‘₯ with respect to πœƒ.

Now, all we need to do to find our slope function, d𝑦 by dπ‘₯, is to find the derivative of 𝑦 with respect to πœƒ and the derivative of π‘₯ with respect to πœƒ. To do this, we remember three rules. We have the product rule, which tells us the derivative of 𝑓 multiplied by 𝑔 is equal to 𝑓 prime multiplied by 𝑔 plus 𝑔 prime multiplied by 𝑓. We have that the derivative of the cosine function is equal to the negative sine function. And finally, the derivative of the sine function is equal to the cosine function.

So to first calculate the derivative of 𝑦 with respect to πœƒ, we differentiate six plus three cos of πœƒ to get negative three sin πœƒ. And then, we multiply this by the sin of πœƒ. Next, we differentiate sin πœƒ to get the cos of πœƒ. And then finally, this gets multiplied by six plus three cos of πœƒ. We can then do the same to calculate the derivative of π‘₯ with respect to πœƒ. We differentiate six plus three multiplied by the cos of πœƒ to get negative three sin of πœƒ. We then multiply this by the cos of πœƒ. Then we add onto this the derivative of the cos of πœƒ which is negative sin πœƒ. And finally, we multiply this by six plus three cos of πœƒ.

We’re now ready to find the slope of our tangent line when πœƒ is equal to πœ‹. From the chain rule, we have that this is equal to the derivative of 𝑦 with respect to πœƒ evaluated at πœ‹ divided by the derivative of π‘₯ with respect to πœƒ evaluated at πœ‹. So we substitute πœƒ is equal to πœ‹ into our equation for d𝑦 by dπœƒ to get the numerator. Similarly, we substitute πœƒ is equal to πœ‹ into our equation for dπ‘₯ by dπœƒ to get the denominator.

Now, by using the fact that sin of πœ‹ is equal to zero and the cos of πœ‹ is equal to negative one, we can evaluate this expression. Evaluating the numerator gives us zero plus negative three. And if we evaluate the denominator, we see that both of our terms have a factor of zero. So we get zero plus zero. This gives us that our slope is equal to negative three divided by zero, which is not well defined. So we can conclude that the slope is undefined at the point three, πœ‹.

To help us see 𝑦, let’s sketch a graph of our polar equation π‘Ÿ is equal to six plus three multiplied by the cos of πœƒ. So we start by sketching a graph of π‘Ÿ is equal to six plus three multiplied by the cos of πœƒ. And the point we’re interested in is when πœƒ is equal to πœ‹. And we can see if we plot the tangent line that we just get a vertical tangent line. Therefore, since our tangent line is vertical at the point where πœƒ is equal to πœ‹, we can conclude that the slope of the polar equation π‘Ÿ is equal to six plus three multiplied by the cos of πœƒ is undefined at the point three, πœ‹.

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