Video: Finding the Equation of the Tangent to the Curve Defined by Trigonometric Parametric Equations at a Point

Find the equation of the tangent to the curve π‘₯ = 5 sec πœƒ and 𝑦 = 5 tan πœƒ at πœƒ = πœ‹/6.

03:30

Video Transcript

Find the equation of the tangent to the curve π‘₯ equals five sec πœƒ and 𝑦 equals five tan πœƒ at πœƒ equals πœ‹ by six.

To find the equation of a tangent, we must begin by calculating its gradient. That, of course, is the value of the derivative at that point. So we’re going to need to begin by calculating the value of d𝑦 by dπ‘₯ when πœƒ is equal to πœ‹ by six.

Well, of course, these are parametric equations. There’s an equation for π‘₯ in terms of πœƒ and an equation for 𝑦 in terms of πœƒ. We recall that, given two parametric equations β€” π‘₯ is some function in πœƒ and 𝑦 is some other function in πœƒ β€” we can find d𝑦 by dπ‘₯ by dividing d𝑦 by dπœƒ by dπ‘₯ by dπœƒ. So we see we’re going to need to begin by differentiating each of our functions with respect to πœƒ. We’ll use the general result for the derivative of sec of π‘₯ and tan of π‘₯ to do so.

The derivative of sec of π‘₯ is sec π‘₯ tan π‘₯. And the derivative of tan π‘₯ is sec squared π‘₯. This means that dπ‘₯ by dπœƒ is five sec πœƒ tan πœƒ. And then d𝑦 by dπœƒ is five sec squared πœƒ. d𝑦 by dπ‘₯ is the quotient to these. It’s five sec squared πœƒ divided by five sec πœƒ tan πœƒ. And of course, we can simplify by dividing through by five and by sec πœƒ. We then recall that sec πœƒ is the same as one over cos πœƒ. We then recall that sec πœƒ is the same as one over cos πœƒ. And we’re going to be dividing this by tan πœƒ, or dividing by sin πœƒ over cos πœƒ.

Now dividing by a fraction is the same as multiplying by the reciprocal of that fraction. So we’re going to be multiplying one over cos πœƒ by cos πœƒ over sin πœƒ. And that’s great because we have a little more canceling to do. We’ve now fully simplified the expression for the derivative of 𝑦 with respect π‘₯. It’s one over sin πœƒ. We might write that as csc πœƒ.

Remember, we were looking to find the value of the gradient of the tangent at πœƒ equals πœ‹ by six. So let’s work out d𝑦 by dπ‘₯ when πœƒ is πœ‹ by six. That’s one over sin of πœ‹ by six. And since sin of πœ‹ by six is one-half, we find the gradient of the tangent to be two.

Now that we have the gradient to the curve, let’s clear some space and see what else we need. The equation of a straight line with the gradient of π‘š that passes through a point with Cartesian coordinates π‘₯ one, 𝑦 one is 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. We quite clearly know the value of π‘š. But we don’t have an π‘₯𝑦-coordinate we can use.

But by substituting πœƒ equals πœ‹ by six into each of our original equations, we’ll find the coordinate of the point where the tangent meets the curve. We get π‘₯ equals five sec of πœ‹ by six. And actually that’s 10 root three over three. And then we get 𝑦 to be equal to five tan of πœ‹ by six. And that’s five root three over three.

Substituting what we have into our formula for the equation of a straight line, and we get 𝑦 minus five root three over three equals two times π‘₯ minus 10 root three over three. When we distribute the parentheses, we get two π‘₯ minus 20 root three over three on the right-hand side. We rearrange by subtracting two π‘₯ and adding 20 root three over three to both sides to find the equation to be 𝑦 minus two π‘₯ plus 15 root three over three equals zero.

All that’s left is for us to simplify our fraction. And when we do, we find the equation of the tangent to our curve when πœƒ is equal to πœ‹ by six to be 𝑦 minus two π‘₯ plus five root three equals zero.

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