79–82. {Use of Tech} Visualizing tangent and normal lines

Question

a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)

x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

Accepted Answer

Hello. In this video, we are going to be determined the equations of the tangent line and normal line to the following curve at the given point. Now, the curve given to us is X4th power equal to 1/8 multiplied by the quantity X2 + Y2. And we want to determine the equations of both of these lines at the 0.2 negative. So how do we start this problem? Well, let's first start by finding the equation to the tangent line. Now, in order to find the equation to the tangent line, there's two pieces of information we need. First, we need a defined point, which is given to us as negative half, negative half, but then we need to find the slope of the tangent line. In order to find the slope of the tangent line, we are going to need to take the derivative of the given curve. Before we take the derivative of the given curve, let's rewrite the curve by distributing 1/8 to X2 plus Y2. That is going to give us the equation as X fourth power, equal to 1/8 multiplied by X2 + 1/8 multiplied by Y2. Now what we are going to do is we are going to take the derivative of this curve with respect to X. On the left hand side, the derivative of X the 4th power is going to be 4 multiplied by X to the 3rd power by the power rule. On the right hand side, we will take 1/8 and multiply this by the derivative of X2, which is going to be 2 X by the power rule. Then, for the second term, we will take 1/8 and multiply this by the derivative of Y2. That is going to be 2 Y by the power rule, but because we are taking the derivative of Y with respect to X, that is going to generate a DYDX term through implicit differentiation. Now what we want to do is simplify. We will have 4 X cubed. Equal to 1/8 multiplied by 2 X, which is going to simplify to be 1/4 X. Then 1/8 multiplied by 2 YDYDX is going to simplify to be 1/4 YDYDX. Now what we want to do is we want to solve for the derivative DYDX. We will go ahead and start by subtracting 1/4 X to both sides of the equation and reordering the terms. That is going to give us 1/4 YYDX equal to 4 X cubed minus 1/4 X. Now let's go ahead and multiply 14th and Y together. That is going to give us Y divided by 4. Now, in order to isolate DYDX, we are going to have to multiply both sides of this equation by the reciprocal of Y divided by 4. That is going to be 4 divided by Y. And that will allow us to rewrite the derivative as D Y divided by DX equal to 4 divided by Y multiplied by the quantity 4 X cubed. Minus 1/4 X. Now in order to obtain the slope to the tangent line, we are going to plug in the given values of the point. That point is 1/2. That is going to simplify the derivative to be 4 divided by -12 multiplied by the quantity, 4 multiplied by -12 cubed minus 1/4 multiplied by 1/2. This is going to simplify to be -8. Multiplied by the quantity. 4 multiplied by negative, 1 divided by 8, plus 1 divided by 8. 4 multiplied by -18 will be 12, but to keep everything as the same denominator, we will leave this as -4 divided by 8. This will simplify to be -8, multiplied by -4 divided by 8 plus 1/8. This will further simplify to be -8, multiplied by the quantity, -3 divided by 8, and this will finally simplify to be 3. So what this means is that the slope of the tangent line is going to be 3. Since we have the slope to the tangent line and the defined point, negative half, common negative half, we can use these two pieces of information to find the equation to the tangent line. So For the tangent line. We will use the point slope form Y minus Y1 equal to the slope M multiplied by X minus X1. That is going to be Y minus -5, which will be equal to 3. Multiplied by X minus -2, which will be positive. We will go ahead and distribute 3 across X + 1/2. That will give us Y equal to, sorry, Y + 1/2. Equal to 3 X plus 3 divided by 2. Finally, if we subtract one half to both sides of the equation, that is going to give us Y is equal to 3 X plus 2 divided by 2, which will simplify to be 1. This is going to be the equation to the tangent line. Now what we need to do is we need to solve for the second half of this problem and find the equation to the normal line. Now, the equation of the normal line is similar to the equation of the tangent line, but the normal line is a line that is perpendicular to the tangent line. So what that means is that in order to solve for the slope of the normal line, we will need to take the negative reciprocal of the tangent line. That is going to be negative 1 divided by 3, and this is going to be the slope that we will use for the normal line. Now, repeating the process by using the point slope form, and now the slope of the normal line, we will go ahead and solve for the equation of the normal line. That is going to be Y minus -2, which is equal to the slope -13, multiplied by the quantity X minus -2, which will be positive. We will distribute 1/3 to X + 1/2, and that will give us Y + 1/2 equal to -13 X minus 16. And if we subtract 1/2 to both sides of the equation, that will give us Y is equal to -13 X. Minus 2/3, and this is going to be the equation to the normal line. So again The equation to the tangent line will be Y equal to 3 x + 1, and by taking the negative reciprocal of the slope to the tangent line, we were able to solve for the equation of the normal line, which is Y equal to -13 X minus 2/3. So, I hope this video helps you in understanding how to approach this type of problem, and I will go ahead and see you all in the next video.

79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

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79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

Watch next

79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)