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

Question

b. Graph the tangent and normal lines on the given graph.

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

Accepted Answer

Hello. In this video, we are going to be graphing the tangent line and normal line to the following curve at the given point. Now, the curve given to us is 2 X to the fourth power, equal to 81 multiplied by X2 plus Y2. Now, we want to find the equations of the tangent line and the normal line passing through the 0.9 9. So, how do we start this problem? Well, since we are finding the equations of the tangent line and the normal line, we will first need to start by finding the equation to the tangent line. The reason why we need to find the equation to the tangent line first is because we first need to obtain the equation to the slope of the tangent line in order to find the slope of the normal line. So, let's go ahead and start by finding the equation to the tangent line. Now, in order to find the equation to the tangent line, we first need to calculate the slope of the tangent line. In order to do this, we can take the derivative of the given curve, plug in the given 0.9.9, and that will be the slope of the tangent line passing through the 0.9.9. So what we will do is we will take the derivative with respect to X of the given curve to X to the power of 4 equal to 81, multiplied by the quantity X2 plus Y2. Let's go ahead and take the derivative with respect to X. Before we take the derivative, we will go ahead and distribute 81 into the quantity X2 + Y2. That will allow us to rewrite the derivative as to X the power of 4, equal to 81 X2 plus 81 Y2. Now let's go ahead and take the derivative. On the left hand side, 2 X to the power of 4, that derivative will be 8 X cubed through the power rule. On the right-hand side, the derivative of 81 X squad will be 162 X through the power rule. And the derivative of 81 Y squad will be 162 Y through the power rule, but we need to be careful because since we are taking a derivative of Y with respect to X, we will generate a DYDX term through implicit differentiation. Now, our goal here is to solve for the derivative DYDX. We will start by subtracting both sides of the equation by 162 X. That will allow us to rewrite the derivative as 162 Y D YDX is equal to 8 X cubed minus 162 X. Finally, in order to solve for DYDX, we will divide both sides of the equation by 162 Y. That will leave us with DYDX equal to 8 X cubed minus 162 X divided by 162 Y. Now that we have the derivative, let's go ahead and plug in the point 9.9 in order to find the equation to the tangent line. When we plug in the value 9.9 for the tangent line. That will give us a slope that is 8, multiplied by 9 cubed minus 162 multiplied by 9, all divided by 162 multiplied by 9. This will simplify to give us the value 4,374 divided by 1,458. And this will further simplify to give us the value of 3. So the slope to the tangent line is going to be 3. Now that we have the slope to the tangent line, we can find the equation to the tangent line by using the slope and the given 09 9 in the point slope form. That will be Y minus 9 equal to the slope 3 multiplied by X minus 9. We will go ahead and distribute 3 into the quantity X minus 9, leaving us with Y minus 9 equal to 3X minus 27. And if we add both sides of the equation by 9, we will go ahead and get an equation to the tangent line, Y is equal to 3X minus 18. This is going to be the equation to the tangent line. Now, what about the equation of the normal line? Well, first, we need to calculate the slope of the normal line. Now, the slope of the normal line is going to be the negative reciprocal to that of the tangent line. That is going to give us a slope of negative 1 divided by 3. By using the same 0.99 we will get the point slope form as Y minus 9 equal to -13, multiplied by X minus 9. We will distribute 1/3 to X minus 9, allowing us to rewrite the equation as Y minus 9, equal to -13 X plus 3. Then, by adding 9 to both sides of the equation, we will get an equation of the normal line as Y is equal to -1/3 X plus 12. This is going to be the equation to the normal line. Now that we have the equation of the tangent line and the equation of the normal line, let's go ahead and plot them in the graph above. Now again, we will go ahead and plot the tangent line in red. The equation to the tangy line is Y is equal to 3X minus 18. This equation to the tangent line will pass to the 0.99 and it has a wide intercept of -18, roughly located here. So, I will do my best to draw it, but that equation is roughly gonna look like this. It is a straight line that touches the 0.99 and passes the point. 0.18. Now we will go ahead and graph the equation of the normal line in black. Recall that the equation of the normal line is Y is equal to -1/3 X plus 12. What this means is that this has a white intercept at 0.12. And again, the equation of the normal line is perpendicular to that of the tangent line. So it is going to pass through the 0.99 but be perpendicular to the equation of the tangent line. That graph of the normal line will roughly look like this. And this is going to be the solution to the problem. So I hope this video helps you in understanding how to graph the equation to the tangent and normal line, and I will go ahead and see you all in the next video.

79–82. {Use of Tech} Visualizing tangent and normal linesb. Graph the tangent and normal lines on the given graph.x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

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79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)