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

Find the slope of the tangent line to π‘Ÿ = πœƒ at πœƒ = πœ‹/2.

01:56

Video Transcript

Find the slope of the tangent line to π‘Ÿ equals πœƒ at πœƒ equals πœ‹ by two.

We’ve been given an equation to a polar curve, π‘Ÿ is equal to some function of πœƒ. And we’re asked to find the slope of the tangent line to the curve at a given point, where πœƒ is equal to πœ‹ by two. To find the slope, we need to find the derivative of 𝑦 with respect to π‘₯. And so, we use a special formula. This formula says that d𝑦 by dπ‘₯ is equal to dπ‘Ÿ by dπœƒ sin πœƒ plus π‘Ÿ cos πœƒ all over dπ‘Ÿ by dπœƒ cos πœƒ minus π‘Ÿ sin πœƒ. We can see we’re going to need to begin then by working out dπ‘Ÿ by dπœƒ. π‘Ÿ is equal to πœƒ, and the derivative of πœƒ with respect to πœƒ is simply one. So we say that dπ‘Ÿ by dπœƒ is equal to one. Replacing dπ‘Ÿ by dπœƒ in our equation for the slope, we get one sin πœƒ plus π‘Ÿ cos πœƒ over one cos πœƒ minus π‘Ÿ sin πœƒ. And, of course, we don’t really need these ones.

But we’re also told that π‘Ÿ is equal to πœƒ. So let’s replace π‘Ÿ with πœƒ and we have an equation for d𝑦 by dπ‘₯ purely in terms of one variable, in terms of πœƒ. The slope of the tangent line to π‘Ÿ equals πœƒ is given by the value of d𝑦 by dπ‘₯ at the point where πœƒ is equal to πœ‹ by two. So we substitute πœƒ equals πœ‹ by two into sin πœƒ plus πœƒ cos πœƒ over cos πœƒ minus πœƒ sin πœƒ. Now, of course, sin of πœ‹ by two is one. Cos of πœ‹ by two is zero. And we repeat these values on the denominator of our fraction. So we get one plus zero as our numerator. And our denominator is zero minus πœ‹ by two, which simplifies to one over or divided by negative πœ‹ by two.

Now, of course, to divide by a fraction, we multiply by the reciprocal of that fraction. So d𝑦 by dπ‘₯ evaluated at πœƒ equals πœ‹ by two is one multiplied by negative two over πœ‹, which is simply negative two over πœ‹. And so, the slope of the tangent line to π‘Ÿ equals πœƒ at πœƒ equals πœ‹ by two is negative two over πœ‹.

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