60–62. {Use of Tech} Multiple tangent lines Complete the follo...

Question

60–62. {Use of Tech} Multiple tangent lines Complete the following steps.

b. Graph the tangent lines on the given graph.

4x³ =y²(4−x); x=2 (cissoid of Diocles)

Accepted Answer

Hello. In this video, we are going to be sketching the graph of the given function and its tangent lines. We are asked to sketch the graph of the function X cubed equal to Y2 multiplied by 10 minus X, and we want to find its tangent lines at X equal to 5. We will then graph these equations by using a graphing utility. Now, since we are using a graphing utility, we don't have to solve for any points for the given curve. We can just type in the curve into whatever graphing utility we choose. However, what we do need to do is we need to solve for the equations of the tangent lines that pass through the value X is equal to 5. So, in order to obtain tangent lines, there are two values that we need. First, we need to obtain the relative points X Y. These are going to be the points that the tangent lines pass through. Next, we will need to solve for the slope of the tangent lines. Let's go ahead and start by solving for the slope of the tangent lines. In order to obtain the slopes of the tangent line, we will need to take the derivative of the given curve. Now, we will be taking the derivative with respect to X. This will be the derivative of X cubed equal to Y2 multiplied by 10 minus X. Now, before we take the derivative, let's go ahead and distribute Y2 into the quantity 10 minus X. This will give us the derivative with respect to X of X cubed equal to 10 Y2 minus XY2. Now let's go ahead and take the derivative. The derivative of X cubed with respect to X will be 3X2 through the power rule. The derivative of 10 Y squad will be 20 Y through the power rule. However, since we are taking a derivative of Y with respect to X, we will generate a DYDX term through implicit differentiation. Then, in order to take the derivative of XY2, we will need to use the product rule. We will subtract the product rule of XY2d. That will be X, multiplied by the derivative of Y, which will be 2 Y through the power rule, but again, this is a derivative of Y with respect to X, so we will get a DYDX term. Then we will add Y2 multiplied by the derivative of X, which is going to be 1 through the power rule. Now what we will do is we will simplify. If we distribute -1 into the product rule of XY2d, that will allow us to rewrite the derivative as 3 X2 equal to 20 DYDX minus 2 XYDYDX minus Y2. Now, what we want to do is we want to keep all the terms with DYDX on the right-hand side of the equation, and we are going to move any term that does not have DYDX to the left-hand side of the equation. This will allow us to rewrite the derivative as 3 X2 plus Y2 equal to 20 DYDX minus 2XYDYDX. Now, since we are solving for the derivative DYDX, what we can go ahead and do is we can factor a DYDX from the right-hand side of the equation. That will allow us to rewrite the derivative S3X2 plus Y2 equal to the quantity 20 Y minus 2 XY multiplied by DYDX. Finally, in order to solve for the derivative DYDX, we will divide both sides of the equation by 20 Y minus 2XY. That will leave us with DYDX equal to 3 X 2 plus Y2 divided by 20 Y minus 2 XY. This is the derivative of the given curve, and this will be the derivative we use to solve the slope of the tangent line. Now that we have the equation that will give us the slope of the tangent line, let's go ahead and solve for the respective point X. Y. Now, we are already told that the X value that the tangent line passes through is X is equal to 5. In order to find the respective Y value, we will take this value of X and plug it in to the original curve that was given to us. Now, again, that curve is X cubed. Equal to Y2 multiplied by 10 minus X. By plugging in the value of 5, we can rewrite the curve as 5 cubed equal to Y2 multiplied by 10 minus 5. Now, let's go ahead and simplify. 5 cubed will simplify to be 125. 10 minus 5 will be 5, and 5 multiplied by Y2 will give us 5 Y squared. We will divide both sides of this equation by 5, and that will leave us with Y2 equal to 25. Finally, if we take the square root of both sides of this equation, we will end up with two values of Y. Y is equal to positive and negative 5. What this means is that we get 2 points at the value X is equal to 5. We have the 1st. 5.5, and we have the second point 5.5. What this means is that there are going to be two equations for tangent lines to the given curve. What we need to do is we need to solve for the slope at both points of the tangent lines, using the derivative we solved for earlier. Let's go ahead and solve for the equation of the tangent line at the point 5.5. By taking this value and plugging it into the derivative that we saw for earlier, the slope at 5.5 will equal to 3 multiplied by 5 squad. Plus 5 squared, divided by the quantity, 20 multiplied by 5, minus 2 multiplied by 5, multiplied by 5. This will simplify to leave us with 100 divided by 50, with which further simplifies to be 2. What this means is that the tangy line passing through the 0. 5.5, will have a slope of 2. Now we are going to plug this into the point slope form in order to solve for the equation of the tangent line. That is going to be Y minus Y1, which is 5, equal to the slope 2 multiplied by X minus X1, which is 5. We will distribute 2 across X minus 5, which will give us Y minus 5, equal to 2 X minus 10. And adding 5 to both sides of this equation will leave us with Y is equal to 2X minus 5. This is going to be the equation of the tangent line passing through the point 5.5. Now we are going to repeat this process, but with the 05-5. By plugging in these values into the derivative we solve for, the slope of the tangy line passing through this point will be 3 multiplied by 5 squared. Plus 5 squad divided by 20. Multiply by -5, -2, multiplied by 5, multiply by -5. This will simplify to be 100 divided by -50, which further simplifies to be -2. So this tangent line will have a slope of -2. Plugging it into the point slope form will give us Y minus -5, which is 5, equal to -2 multiplied by X minus 5. Distributing -2 to X minus 5 will give us Y + 5 equal to -2 X + 10. And if we subtract 5 to both sides of the equation, we get Y is equal to -2 X plus 5. This is going to be the equation of the second tangent line. Now that we have the equations of both tangent lines and the original curve, we will now plot them all using a graphing utility. So what I have here is the plot of the tangent lines and the giving curve. The given curve is the curve in black, that is X squared. E X cubed equal to Y2 multiplied by 10 minus X. The red line is the equation to the tangent line passing through the 0.5.5, and the blue line is the equation of the tangent line passing to the point 5.5. And with that being said, that is the solution to this problem. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
b. Graph the tangent lines on the given graph.
4x³ =y²(4−x); x=2 (cissoid of Diocles)

Key Concepts

Tangent Lines

Derivatives

Graphing Techniques

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