Video: Finding the Slope of the Tangent to a Polar Curve at a Given Theta Value

Find the slope of the tangent line to the polar curve π‘Ÿ = 1 + cos πœƒ at the point πœƒ = πœ‹/4.

07:10

Video Transcript

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

We’re given a curve defined by a polar equation. And we need to find the slope of the tangent line to this polar 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 a tangent line is given by d𝑦 by dπ‘₯, the rate of change of 𝑦 with respect to π‘₯. Normally, when we’re finding this, we’re given 𝑦 as a function in π‘₯, however, this time we’re given a polar curve. So we’re going to need to recall what we know about polar curves to find d𝑦 by dπ‘₯ in this case.

We recall by using the chain rule and the inverse function theorem, we got the following equation for the slope of a polar curve in terms of d𝑦 by dπ‘₯. We get that d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. And this will only be true so long as our denominator dπ‘₯ by dπœƒ is not equal to zero. So to find an expression for d𝑦 by dπ‘₯, we’re first going to need to find expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ. And to find expressions for these, we’re going to need to write 𝑦 in terms of πœƒ and π‘₯ in terms of πœƒ. And we can do this by recalling our general equations for polar coordinates. 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.

Remember, for this polar curve, we’re told that π‘Ÿ is equal to one plus the cos of πœƒ. So we can just substitute this into our equations. Substituting in π‘Ÿ is equal to one plus the cos of πœƒ, we get that 𝑦 is equal to one plus the cos of πœƒ all multiplied by the sin of πœƒ and π‘₯ is equal to one plus the cos of πœƒ all multiplied by the cos of πœƒ. So we’ve now written 𝑦 as a function of πœƒ and π‘₯ as a function of πœƒ. So we can differentiate these with respect to πœƒ.

Let’s start with finding d𝑦 by dπœƒ. We’ll start by distributing the sin of πœƒ over our parentheses. Doing this, we get that 𝑦 is equal to the sin of πœƒ plus the cos of πœƒ times the sin of πœƒ. Now, to find an expression for d𝑦 by dπœƒ, we need to differentiate this term by term. Let’s start with differentiating the sin of πœƒ with respect to πœƒ. And this is a standard trigonometric derivative result we should know. The derivative of the sin of πœƒ with respect to πœƒ is the cos of πœƒ.

Now, there are several different ways of differentiating our second term of the cos of πœƒ multiplied by the sin of πœƒ. For example, we could do this by using the double-angle formula for sin. However, since this is the product of two differentiable functions, we’ll do this by using the product rule. We recall the product rule tells us the derivative of the product of two differentiable functions 𝑒 of π‘₯ times 𝑣 of π‘₯ with respect to π‘₯ is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ plus 𝑣 prime of π‘₯ times 𝑒 of π‘₯. And in our case, our function is in terms of πœƒ and we’re differentiating in terms of πœƒ. So we’ll rewrite this in terms of πœƒ.

So to use the product rule, we’ll set our function 𝑒 of πœƒ to be the cos of πœƒ and 𝑣 of πœƒ to be the sin of πœƒ. To use the product rule, we need to find expressions for 𝑒 prime of πœƒ and 𝑣 prime of πœƒ. Both of these are standard trigonometric derivative results which we should commit to memory. We get 𝑒 prime of πœƒ is negative the sin of πœƒ and 𝑣 prime of πœƒ is the cos of πœƒ. We’re now ready to apply the product rule to differentiate the cos of πœƒ times the sin of πœƒ with respect to πœƒ. We just need to substitute our expressions for 𝑒 of πœƒ, 𝑣 of πœƒ, 𝑒 prime of πœƒ, and 𝑣 prime of πœƒ into our formula for the product rule.

We get negative the sin of πœƒ times the sin of πœƒ plus the cos of πœƒ multiplied by the cos of πœƒ. And of course, we can simplify this to give us the cos squared of πœƒ minus the sin squared of πœƒ. Now, we can use this formula to find our expression for d𝑦 by dπœƒ. We get d𝑦 by dπœƒ is equal to the cos of πœƒ plus the cos squared of πœƒ minus the sin squared of πœƒ. We now want to find an expression for dπ‘₯ by dπœƒ. This time, we’ll start by distributing the cos of πœƒ over our parentheses. We get the cos of πœƒ plus the cos squared of πœƒ.

We can now differentiate this term by term. First, the derivative of the cos of πœƒ with respect to πœƒ is a standard trigonometric derivative result we should commit to memory. It’s negative the sin of πœƒ. Now, there are several different methods we could use to evaluate the derivative of the cos squared of πœƒ. For example, we could rewrite this by using the double-angle formula for cos. We could also do this by using the product rule or the chain rule. Any of these methods would work, and it’s personal preference which one we would use.

So let’s clear some space so we can evaluate this derivative. We’ll do this by using a variance of the chain rule called the general power rule. We recall this tells us for differentiable function 𝑔 of πœƒ and any constant value of 𝑛 the derivative of 𝑔 of πœƒ all raised to the power of 𝑛 with respect to πœƒ is equal to 𝑛 times 𝑔 prime of πœƒ multiplied by 𝑔 of πœƒ raised to the power of 𝑛 minus one. We want to use this to differentiate the cos squared of πœƒ. And to do this, we need to notice the cos squared of πœƒ is just the cos of πœƒ all squared. So to apply the general power rule, we set our exponent 𝑛 equal to two and 𝑔 of πœƒ to be the cos of πœƒ.

To use the general power rule, we need an expression for 𝑔 prime of πœƒ. That’s the derivative of the cos of πœƒ with respect to πœƒ, and we know this is negative the sin of πœƒ. Substituting 𝑛 is equal to two in our expressions for 𝑔 of πœƒ and 𝑔 prime of πœƒ into the general power rule, we get the derivative of the cos squared of πœƒ with respect to πœƒ is equal to two times negative the sin of πœƒ multiplied by the cos of πœƒ all raised to the power of two minus one. And we can then simplify this. First, in our exponent, two minus one simplifies to give us one, and then raising our function to the power of one doesn’t change the function. Finally, we then multiply these terms together to get negative two sin πœƒ times the cos of πœƒ.

We’re now ready to use this to find an expression for dπ‘₯ by dπœƒ. We get dπ‘₯ by dπœƒ is equal to negative the sin of πœƒ minus two times the sin of πœƒ times the cos of πœƒ. And now that we found expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ, we can use our formula to find an expression for d𝑦 by dπ‘₯. We just need to substitute our expressions for d𝑦 by dπœƒ and dπ‘₯ by dπœƒ into the formula. Doing this, we get d𝑦 by dπ‘₯ is equal to the cos of πœƒ plus the cos squared of πœƒ minus the sin squared of πœƒ all divided by negative the sin of πœƒ minus two times the sin of πœƒ multiplied by the cos of πœƒ.

But remember, the question is not asking us to find an expression for d𝑦 by dπ‘₯. Instead, we’re asked to find the slope of our tangent line when πœƒ is equal to πœ‹ by four. So we just need to substitute πœƒ is equal to πœ‹ by four into this expression. Substituting πœƒ is equal to πœ‹ by four into this expression, we get the slope of the tangent line to our polar curve at the point where πœƒ is equal to πœ‹ by four is given by the cos of πœ‹ by four plus the cos squared of πœ‹ by four minus the sin squared of πœ‹ by four all divided by negative the sin of πœ‹ by four minus two times the sin of πœ‹ by four multiplied by the cos of πœ‹ by four.

And we could answer this by using what we know about trigonometric functions evaluated at standard angles in radians. However, we can also do this by using our calculator. In either method, we get negative root two plus one. And this is our final answer.

Therefore, by using our formula for finding d𝑦 by dπ‘₯ for a polar curve, we were able to show the slope of the tangent line to the polar curve π‘Ÿ is equal to one plus the cos of πœƒ at the point where πœƒ is equal to πœ‹ by four is equal to negative root two plus one.

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