73–78. {Use of Tech} Normal lines A normal line at a point P o...

Question

73–78. {Use of Tech} Normal lines A normal line at a point P on a curve passes through P and is perpendicular to the line tangent to the curve at P (see figure). Use the following equations and graphs to determine an equation of the normal line at the given point. Illustrate your work by graphing the curve with the normal line. <IMAGE>

Exercise 46

Exercise 46

Accepted Answer

Hello. In this video, we are going to be determining the equation to the normal line of the following curve at the given point. Now, the curve given to us is X cubed plus Y cubed equal to 8 XY. And we want to find the equation of the normal line at the point 4.4. In order to find the equation of the normal line, we first need to find the slope of the tangent line. Now, in order to find the slope of the tangent line, we first need to take the derivative of the given equation. In order to do this, we will take the derivative with respect to X. So, we will need to take the derivative using implicit differentiation because we are taking the derivative of a function in terms of X and Y. We will start by taking the derivative of the left hand side. The derivative of X cubed will be 3 X 2 by the power rule. And the derivative of Y cubed will be 3 Y squared by the power rule. But since we are taking the derivative of Y with respect to X, that will generate a DYDX term through implicit differentiation. Now we will take the derivative of the right hand side of the equation. That will be 8, multiplied by the product rule of X and Y. That will be X multiplied by the derivative of Y, which will be 1. And again, this will generate a DYDX term since we are taking the derivative of Y with respect to X. Then we will add Y multiplied by the derivative of X, which is going to be 1. So now our goal is we want to solve for the derivative DYDX. We want to move all the terms with the DYDX factor to the left-hand side of the equation. But first we need to simplify. We are going to distribute 8 to XDYDX plus Y. That will allow us to rewrite the equation as 3 X2 + 3 Y2 DYDX equal to 8 XDYDX plus 8 Y. Next, what we are going to do is we are going to move the DYDX terms to the left hand side of the equation, and the constant terms to the right hand side. That will allow us to rewrite the equation as 3 Y 2 D YDX minus 8 XDYDX equal to 8 Y minus 3X 2. Since we are solving for DYDX we can factor a DYDX from the left-hand side of the equation. That will give us DYDX multiplied by the quantity 3 Y2 minus 8X equal to 8 Y minus 3 X 2. And in order to solve for DYDX, we will divide both sides of this equation by the quantity 3 Y2 minus 8X. That will give us a simplified derivative DYDX equal to 8 Y minus 3X 2 divided by 3 Y2 minus 8 X. This is going to be the simplest form of the derivative, and now we want to find the slope of the tangent line at the point 4.4. In order to do this, we will go ahead and plug in the values of 4.4 to the left hand side of the equation. That will go ahead and give us the slope or the derivative as 8 multiplied by 4, minus 3 multiplied by 4 squared, divided by 3 multiplied by 4 squad, minus 8 multiplied by 4. This will simplify to be negative 16 divided by 16, which is going to give us -1. That means that the slope of the tangent line is going to equal to -1. But recall that we are solving for the equation of the normal line, not the tangent line. The normal line is a line that is perpendicular. To the tangent line. And since the normal line is perpendicular to the tangent line, we are going to have to take the negative reciprocal of the tangent line slope in order to find the slope of the normal line. The negative reciprocal. Of the tangent line is going to be -1 divided by -1, and this is going to simplify to be positive 1. So the slope of the normal line will be 1. Since we have the slope, and we have the 0.4.4, we can use the point slope form to find the equation of the normal line. That will be Y minus 4 equal to the slope 1 multiplied by the quantity X minus 4. This is going to simplify the equation to be Y minus 4 equal to X minus 4, and if we add 4 to both sides of the equation, this will simplify to be Y is equal to X. This is going to be the equation to the normal line at the 0 4.4. And with that being said, the solution to this problem is going to be D. So, I hope this video helps you in understanding on how to find the equation to the normal line, and I will go ahead and see you all in the next video.

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