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 + sin πœƒ at πœƒ = πœ‹/4.

06:21

Video Transcript

Find the slope of the tangent line to the curve π‘Ÿ is equal to one plus the sin of πœƒ at πœƒ is equal to πœ‹ by four.

We’re given a curve defined by the polar equation π‘Ÿ is equal to one plus the sin of πœƒ. We need to use this to determine the slope of the tangent line to this curve at the point where πœƒ is equal to πœ‹ by four. To do this, we first need to recall that the slope of the tangent line to a curve is given by d𝑦 by dπ‘₯, the rate of change in 𝑦 with respect to π‘₯. And we need to find this value when πœƒ is equal to πœ‹ by four. However, in this case, we’re not given 𝑦 as a function in π‘₯. We can’t evaluate this directly. Instead, we’re given a polar equation. π‘Ÿ is equal to one plus the sin of πœƒ. Instead, we’re going to need to use what we know about polar equations.

First, let’s recall the general polar equations. 𝑦 is equal to π‘Ÿ times the sin of πœƒ, and π‘₯ is equal to π‘Ÿ times the cos of πœƒ. This gives us an equation for 𝑦 in terms of π‘Ÿ and πœƒ and an equation for π‘₯ in terms of π‘Ÿ and πœƒ. However, we have an equation for this curve for π‘Ÿ in terms of πœƒ. So if we substitute this equation in, we’ll have equations for π‘₯ in terms of πœƒ and 𝑦 in terms of πœƒ. Substituting in π‘Ÿ is equal to one plus the sin of πœƒ, we get 𝑦 is equal to one plus the sin of πœƒ all multiplied by the sin of πœƒ. And π‘₯ is equal to one plus the sin of πœƒ all multiplied by the cos of πœƒ.

We now have 𝑦 in terms of the variable πœƒ and π‘₯ in terms of the variable πœƒ. We want to use these to find an expression for d𝑦 by dπ‘₯. And luckily, we know how to do this by using the chain rule. So by either applying the chain rule or by using what we know about the slope of polar curves, we get d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. And this will be valid so long as dπ‘₯ by dπœƒ is not equal to zero. So to find an expression for d𝑦 by dπ‘₯, we just need to find expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ.

We’ll start with d𝑦 by dπœƒ. That’s the derivative of 𝑦 with respect to πœƒ. We’ll start by distributing the sin of πœƒ over our parentheses. This gives us the sin of πœƒ plus the sin squared of πœƒ. We’re going to want to differentiate this with respect to πœƒ. And there’s several different ways of doing this. For example, we could differentiate the sin squared of πœƒ by using the product rule, the chain rule, or the general power rule. And all of these methods are valid, would work, and would give us the correct answer.

However, there’s one more method we could use. We need to recall the following double-angle identity for the cos of two πœƒ. It’s equivalent to one minus two times the sin squared of πœƒ. We want to rearrange this identity to make the sin squared of πœƒ the subject. So we’ll subtract one from both sides of this equivalent and then divide through by negative two. We get one minus the cos of two πœƒ all divided by two is equivalent to the sin squared of πœƒ. And we can then see this is a much easier expression to differentiate with respect to πœƒ.

So, by using the double-angle formula for cos to rewrite the sin squared, we get 𝑦 is equal to the sin of πœƒ plus one minus the cos of πœƒ all divided by two. We’re now ready to find an expression for d𝑦 by dπœƒ. We need to differentiate our expression for 𝑦 with respect to πœƒ term by term. First, we know the derivative of the sin of πœƒ with respect to πœƒ is equal to the cos of πœƒ. Next, we know the derivative of the constant one-half with respect to πœƒ is equal to zero. So we don’t need to include this in our expression.

Now, all we need to do is to differentiate our final term. We know the derivative of the cos of two πœƒ with respect to πœƒ is negative two times the sin of two πœƒ. Then all we need to do is simplify this expression. We have two divided by two, which simplifies to give us one. And then negative one multiplied by negative one simplifies to give us one. So we simplified our expression to show d𝑦 by dπœƒ is equal to the cos of πœƒ plus the sin of two πœƒ.

We now want to do the same to find an expression for dπ‘₯ by dπœƒ. We’ll start by distributing the cos of πœƒ over our parentheses. We get π‘₯ is equal to the cos of πœƒ plus the sin of πœƒ multiplied by the cos of πœƒ. And we’re going to want to differentiate this with respect to πœƒ. Once again, there’s several different methods we could use to differentiate the sin of πœƒ multiplied by the cos of πœƒ with respect to πœƒ. For example, we could do this by using the product rule.

However, this is not the only way of evaluating this derivative. We can also use the double-angle formula for sin. Recall the sin of two πœƒ is equivalent to two times the sin of πœƒ multiplied by the cos of πœƒ. And if we divide this equivalent through by two, we see that one-half sin of two πœƒ is equivalent to sin of πœƒ times the cos of πœƒ. And one-half times the sin of two πœƒ is much easier to differentiate with respect to πœƒ. So we’ll substitute this into our equation for π‘₯. We get π‘₯ is equal to the cos of πœƒ plus one-half times the sin of two πœƒ.

We’re now ready to find an expression for dπ‘₯ by dπœƒ by differentiating our expression for π‘₯ term by term. First, the derivative of the cos of πœƒ with respect to πœƒ is equal to negative the sin of πœƒ. Next, the derivative of one-half sin of two πœƒ with respect to πœƒ will give us two times the cos of two πœƒ divided by two. And we can simplify this by canceling the shared factor of two in our numerator and our denominator. So we’ve shown dπ‘₯ by dπœƒ is equal to negative sin of πœƒ plus the cos of two πœƒ. We’re now ready to use our expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ to find an expression for d𝑦 by dπ‘₯.

Substituting in our expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ, we get d𝑦 by dπ‘₯ is equal to the cos of πœƒ plus the sin of two πœƒ all divided by negative the sin of πœƒ plus the cos of two πœƒ. But we’re not done yet. Remember, the question is asking us to find the slope of the tangent line when πœƒ is equal to πœ‹ by four. So we need to substitute πœƒ is equal to πœ‹ by four into this expression.

Substituting in πœƒ is equal to πœ‹ by four, we get the slope of our tangent line when πœƒ is equal to πœ‹ by four is given by the cos of πœ‹ by four plus the sin of two times πœ‹ by four all divided by negative the sin of πœ‹ by four plus the cos of two times πœ‹ by four. And we could evaluate this by using what we know about trigonometric functions evaluated at standard angles. However, if we evaluate this, we will also get negative root two minus one. And this gives us our final answer.

Therefore, given the curve defined by the polar equation π‘Ÿ is equal to one plus the sin of πœƒ by using our formula for the slope d𝑦 by dπ‘₯, we were able to find the slope of the tangent line to this polar curve when πœƒ is equal to πœ‹ by four. We got that this slope was equal to negative root two minus one.

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