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

Question

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

a. Find equations of all lines tangent to the curve at the given value of x.

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

Accepted Answer

Welcome back, everyone. Determine the equations of all lines tangent to 9 X cubed equals Y2 multiplied by 6 minus X at X equals 3. So first of all, if we want to identify the equation of a tension line, we want to identify the slope of that line, and to do that, we essentially want to solve for the derivative of Y. What we can do is simply apply implicit differentiation for the given expression. We're going to take the derivative of the left hand side, so we're taking the derivative of 9 X cubed, and we're setting it equal to the derivative of the right hand side. So the derivative of Y2. Multiplied by 6 minus X. On the left hand side we are simply applying the power rule, so we get 9 multiplied by. 3 x squad that's the derivative of X cubed, right, and we're going to get 27 x 2 on the left. On the right we have to apply the product rule. So we are taking the derivative of the first function, so the derivative of Y2d. Multiplying it by the second function, which is 6 minus X, and then we're adding the first function Y2 multiplied by the derivative of the second function 6 minus X. And now let's valuate each derivative. So 27 x 2 is equal to applying the power rule we're going to get 2 Y. But Y is a function of X, so we also apply the chain rule and multiply by Y. Which is then multiplied by 6 minus X, and the derivative of 6 minus X is -1 based on the power rule. So we're going to take minus Y2 because Y2 multiplied by -1 is negative Y2. And now let's rearrange, we're going to take 27 X2. Let's add Y2 to both sides. And this is equal to 26 minus X. Now solving for Y, we're going to end up with 27 X2 plus Y2. Divided by 2 Y multiplied by 6 minus X, and this is the general expression of the slope of a tangent line. And notice that it's dependent on X and Y. And we're given the point of tension c x not equals 3. So what we're going to do is solve for the Y values using the original expression, right? So if we take the original expression, we can rearrange for Y2d dividing the left hand side by the right hand side. So 9 x cubed is going to be divided by 6 minus X. So if we take square roots of both sides, then y equals positive or negative square root of 9 x cubed. Divided by 6 minus X, there's no necessity to simplify because we're going to look for the possible values. We're going to have two of them. Let's solve for the 1st 1. Y1 is equal to the positive root of now square root of 9 is 3, so we can just put it in front and now we're taking square root of X cubed. Our X is 3, so we get 3 cubed. In our numerator And dividing that by 6 minus X, so 6 minus 3. Now, 27 divided by 3 gives us 9 square root of 9 is 3, and 3 multiplied by 3 is 9. Now, our second solution is basically the negative value of that result. So it's a negative of Y1. Which is -9. And this means that we have two points of tangency. Our first point is if we 9. And the second point is 3.9. So we have two slopes, right? We have to evaluate each slope at each point. So why? Can be equal to 27 multiplied by X2, which is 32 plus Y2 which is 92. Divided by 2 Y, so 2 multiplied by 9, multiplied by 6 minus X, which is 6 minus 3. Our second possible slope is 27 multiplied by 32 + -9 squad. We will still have the same numerator. But for the denominator, we end up with 2 multiplied by -9. Multiplied by 6 minus 3. So let's evaluate each result. For the top one, we're going to end up with 6, and for the bottom one, we're going to get -6. So those are the possible slopes. And now knowing each slope we can use the point slope formula. So essentially we have to recall that Y minus Y1 is equal to the slope M multiplied by X minus X1. Let's use this expression. We're going to take Y minus Y1. Let's focus on the first point of tangency. Our Y coordinator is 9. Our slope M is 6. And we're multiplying that by x minus X1. X1 is 3. Now expanding this expression, we're going to get Y equals. 6 acts -18 + 9, right? Because we're adding 9 to both sides. So minus 18 + 9 gives us -9. And this is our first. Equation of the first tangent line. And now for the second one, we're taking Y minus Y1. Our Y1 is -9, so we're adding 9 because we're subtracting a negative value. Our slope is -6. Our X is the same. So now solving for Y, we're going to get Y equals -6 X. + 18 minus 9, right? So plus 9. And that's it. We have obtained the two tangent lines at X equals 3. Thank you for watching.

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
a. Find equations of all lines tangent to the curve at the given value of x.
4x³ =y²(4−x); x=2 (cissoid of Diocles)

Key Concepts

Tangent Lines

Implicit Differentiation

Cissoid of Diocles

Watch next