Video: Finding the Slope of the Tangent to a Polar Curve at a Certain Point

Find the slope of the tangent line to the curve π‘Ÿ = 1/πœƒ at πœƒ = πœ‹.

02:09

Video Transcript

Find the slope of the tangent line to the curve π‘Ÿ equals one over πœƒ at πœƒ equals πœ‹.

Remember, when finding the slope, we need to evaluate the derivative of that curve at a specific point. The formula we’re interested in for a polar curve of the form π‘Ÿ equals 𝑓 of πœƒ is d𝑦 by dπ‘₯ equals d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. And then, we have specific equations for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ in terms of π‘Ÿ and πœƒ. We can see that our equation is given as π‘Ÿ equals one over πœƒ. So, we’ll begin by working out the value of dπ‘Ÿ dπœƒ.

Let’s start by writing one over πœƒ as πœƒ to the power of negative one. And of course, to differentiate a simple polynomial terms like this one, we multiply by the exponent and then reduce that exponent by one. So, that’s negative one times πœƒ to the power of negative two, which is the same as negative one over πœƒ squared. We’re also given that πœƒ is equal to πœ‹. So, all we need to do is substitute everything we know into our formula. d𝑦 by dπœƒ is equal to negative one over πœƒ squared times sin πœƒ plus one over πœƒ times cos πœƒ.

And then, we obtain dπ‘₯ by dπœƒ to be equal to negative one over πœƒ squared cos πœƒ minus one over πœƒ sin πœƒ. d𝑦 by dπ‘₯ is the quotient of these too. And of course, we could simplify this somewhat by factoring one over πœƒ, for example. However, we’re going to evaluate this when πœƒ is equal to πœ‹, so we may as well go ahead and substitute that in. When we do, we find that the slope is equal to negative one over πœ‹ squared times sin πœ‹ plus one over πœ‹ times cos πœ‹ over negative one over πœ‹ squared cos πœ‹ minus one over πœ‹ sin πœ‹.

And in fact, sin of πœ‹ is zero and cos of πœ‹ is negative one, so this simplifies really nicely to negative one over πœ‹ over one over πœ‹ squared. Due to the nature of dividing fractions, this simplifies further to negative πœ‹ squared over πœ‹, which is simply negative πœ‹. And we found the value of the derivative at πœƒ equals πœ‹. And therefore, the slope of the tangent line to the curve at this point is negative πœ‹.

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