In Exercises 47 and 48, find an equation for

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Question

In Exercises 47 and 48, find an equation for

(a) the tangent line to the curve at P

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Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says determine the equation of the tangent line to the following curve at point A. Below the problem we're given a graph that has the function Y is equal to 4 plus tangent of X minus 2 multiplied by CX plotted on it. We also have a tangent line that intersects this curve Y at the point A and point A is located at 0.2. Now this problem wants us to determine the equation of the tangent line. And in order to do this, we need to determine the slope of the tangent line first. And so we're going to need to take a derivative of our function Y. So we need to calculate Y and that's going to be equal to the derivative with respect to X of the quantity of 4 plus tangent of X. Minus 2 multiplied by sequant of X. In quantity So taking this derivative, this is going to be equal to. For the first term, we'll have the derivative of 4 with respect to X. Here I'm taking a derivative of a constant and recall that a derivative of a constant is equal to 0. So we'll have 0, plus for the second term, we'll have the derivative with respect to X of tangent of X, and recall that this derivative is equal to c2 of x. Then we'll have minus the derivative with respect to X of 2 multiplied by C X. The 2 is a constant so we can move through the derivatives, so we'll have 2 multiplied by the derivative with respect to X of X, and recall for that derivative that we will get sequent of X multiplied by tangent of X. Now recall that the slope of our tangent line is going to be equal to the derivative at the point where our tangent line intersects our curve. So we need to calculate of 0, since our curve or tangent line intersects our curve at the point of 0.2. So, this is going to be equal to. Sequent squared of 0 minus 2 multiplied by sequent of 0 multiplied by tangent of 0. And when we evaluate this expression, this is equal to 1. Since the sequ of 0 is equal to 1 and the tangent of 0 is equal to 0. So this here is going to be the slope of our tangent line. And since we have our slope as well as a point, we can use our point slope form to write the equation for our tangent line. So we call for the point slope form that will have Y minus our particular Y value, which in this case is 2, and that's going to be equal to our slope, which is 1, multiplied by the quantity of X minus our particular X value, which in this case is 0. So we can simplify this expression, we'll have Y minus 2 is equal to X. And then to solve this for Y, we'll just add 2 to both sides of the equation, and so we get Y is equal to x plus 2. And so this is the equation for our tangent line that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

In Exercises 47 and 48, find an equation for

(a) the tangent line to the curve at P

Key Concepts

Derivative

Trigonometric Derivatives

Equation of a Tangent Line

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