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 48

Exercise 48

Accepted Answer

Hello. In this video, we are going to be finding the equation of a normal line. The problem asks us to determine the equation of the normal line to the following curve at the given point. The curve given to us is X minus XY2 plus Y 4th equal to 24, and we want to determine the equation of the normal line at the 0.22. Now, in order to find this equation, there's two pieces of information we need. We need a given point, which again, is two common negative 2. But then we need the slope of the normal line. In order to obtain the slope of the normal line, we first need to find the slope of the tangent line. The reason why we need the tangent line slope is because the normal line is the negative reciprocal to the tangent line. So let's first start by taking the derivative with respect to X of the given equation. By taking the derivative of the left hand side, we will have 4X3 power by the power rule minus the product rule of XY2. That is going to be X multiplied by the derivative of Y2d. That is going to be 2 Y 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. Then we will add Y2 multiplied by the derivative of X, which is going to be won by the power rule. Finally, we will take the derivative of Y to the fourth power, which will be 4 Y cubed through the power rule, but again, we will generate a DYDX term through implicit differentiation. Then this will equal to the derivative of 24, which will be 0. The derivative of any constant is 0. Now, our goal here is to solve for the derivative DYDX. In order to do this, we will need to simplify the left-hand side of the derivative. We will start by distributing -1 across X multiplied by 2 YDYD X and Y squad. That is going to simplify the derivative to be 4 X cubed minus 2 X Y D YDX minus Y2 + 4 Y cubed DYDX equal to 0. In order to solve for DYDX, we are going to keep all the factors, or all the terms with a factor of DYDX on the left hand side of the equation, and move all the terms without a DYDX factor to the right side of the equation. Thou wilt thou wast to rewrite the derivative as 4 y cubed. DYDX minus 2 X Y DYDX and that will equal to Y2 minus 4 X cubed. What we can do now is we can factor out a DYDX from the left hand side of the equation. That will give us DYDX multiplied by the quantity 4 Y cubed minus 2 XY. Equal to Y2 minus 4 X cubed. And finally, in order to solve for DYDX, we will divide both sides of this equation by 4 Y cubed minus 2XY. That will give us a final derivative DYDX equal to Y2 minus 4 X cubed, divided by the quantity for Y cubed minus 2 XY. What we have here is the derivative to the given curve. And now what we want to do is you want to plug in the 0.22 in order to obtain the slope of the tangent line. That is going to allow us to rewrite the derivative as -2222 minus 4 multiplied by 2 cubed, divided by 4 multiplied by -2 cubed minus 2 multiplied by 2 multiplied by -2. This will simplify to be 4. -32, divided by -32 plus 8. This will further simplify to be -28 divided by -24, which is going to give us 7 divided by 6. This is going to be the slope of the tangent line. Now, again, we are trying to determine the equation of the normal line. Since the equation of the normal line has a slope that is perpendicular to the equation of the tangent line, in order to obtain the slope of the normal line, we will need to take the negative reciprocal of the tangent line slope. That negative reciprocal is going to be -6 divided by 7. This is going to be the slope of the normal line. Now that we have the slope of the normal line and the 0.2 common negative 2, we can use these two pieces of information to determine the equation of the normal line. That is going to give us Y minus -2, which is Y + 2, equal to the slope, -6 divided by 7 multiplied by the quantity X minus 2. We will distribute -6 divided by 7 across X minus 2. That is going to give us Y + 2 equal to -6 divided by 7 X plus 12 divided by 7. And if we subtract 2 to both sides of this equation, this will give us Y is equal to -6 divided by 7 X minus 2 divided by 7. And this is going to be the equation to the normal line at the given point 2-2. And with that being said, the overall solution is going to be C. So I hope this video helps you in understanding how to find the equation of a normal line, and I will go ahead and see you all in the next video.

Key Concepts

Tangent Line

Normal Line

Graphing Techniques

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