The folium of Descartes (See Figure 3.27)
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Question

The folium of Descartes (See Figure 3.27)

c. Find the coordinates of the point A in Figure 3.29 where the folium has a vertical tangent line.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says find the coordinate to the point where the curve 2 XU plus 3 Y cubed equal to 18 multiplied by X multiplied by Y has a vertical tangent line, not including the origin. We're given 4 possible choices as our answers. For choice A, we have the point of 2 multiplied by the cube root of 18, 2 multiplied by the. root of 3. For choice B, we have the point of 2 multiplied by the cube root of 92 multiplied by the cube root of 3. For C, we have the point of 2 multiplied by the cube root of 92 multiplied by the cube root of 9. And for D, we have the point of 2 multiplied by the cube root of 18. 2 multiplied by the cube root of 9. So we need to find the coordinates of the points on our curve that have a vertical tangent line. And to do this, we need to find the derivative of Y with respect to X, and we're gonna do that using implicit differentiation. So we need to take the derivative of both sides of our equation, so we'll have the derivative with respect to X. Of the quantity of 2 X cubed plus 3 Y cubed in quantity, and that's going to be equal to the derivative with respect to X of the quantity of 18 X. Multiplied by y in quantity. And so taking the derivatives of both sides here, on the left hand side, for the first term, that derivative is going to be 6X2d. Then we'll have plus for the derivative of the second term, we'll have 9 Y2 multiplied by the derivative of Y with respect to X since we also have to apply our chain rule. And this is going to be equal to, on the right hand side, we'll have to use our product rule. And so, we'll have 18 multiplied by Y plus 18X multiplied by the derivative of Y with respect to X. And so now we need to solve this equation for the derivative of Y with respect to X. So I'm gonna collect all those terms on the left hand side of the equation and put all other terms to the right hand side. So doing that, we will have 9 multiplied by Y2d, multiplied by the derivative of Y with respect to X. And then we'll need to move the term with 18 X multiplied by the derivative of Y with respect to X to the left hand side from the right hand side. So we'll subtract 18X multiplied by the derivative of Y with respect to X from both sides of the equation. So we'll have minus 18X multiplied by the derivative of Y with respect to X. This is going to be equal to 18 Y. Then we'll have to move the term 6X2d from the left hand side to the right hand side by subtracting 6X2d from both sides, but then we'll have minus 6X2d. And now we can factor out a derivative of Y with respect to X out of both terms on the left hand side of the equation. In doing that, we'll have the derivative of Y with respect to X, multiplied by the quantity of 9 Y2 minus 18 X in quantity, and that's going to be equal to 18 Y. Minus 6 x 2. So dividing over the term multiplying the derivative Y with respect to X, we will get the derivative of Y with respect to X. That's going to be equal to the quantity of 18 Y minus 6 X squad in quantity, divided by the quantity of 9 Y2 minus 18 X in quantity. And here we can factor out a 3 out of both the numerator and denominator, and so those will cancel out. And so that means we'll have the derivative of Y with respect to X, and that's going to be equal to the quantity of 6 Y minus 2 X 2 in quantity, divided by the quantity of 3 Y2 minus 6 X in quantity. And we're looking for the points where we have a vertical tangent line. And recall that we'll have a vertical tangent line whenever the derivative of Y with respect to X is undefined. So in our case, that means that the denominator here is going to be equal to 0. So we'll set 3 Y2 minus 6 X equal to 0, and we're going to solve this expression for X. So to do that, we'll add 6 X to both sides of the equation. So we'll have 3 Y2 is equal to 6 X, and divided both sides by 6, we'll get X is equal to Y2 divided by 2. And so now what we're going to do is substitute this expression back into our original equation. So doing that, we'll have 2 multiplied by the quantity of Y2d divided by 2 in quantity, cubed. Plus 3 cubed. And that's going to be equal to 18, multiplied by the quantity of Y2d divided by 2 in quantity multiplied by Y. and we need to solve this expression for Y. So we'll want to move all of our turns over to the left hand side of the equation. And along the way we'll go ahead and simplify our expression. So for our first term, when we simplify it, we're going to get Y 6 divided by 4. Then we'll have + 3 Y cubed. Minus 9 Y cubed, and that's gonna be equal to 0. So, collecting like terms, we're going to have to the 6th divided by 4. Minus 6 y cubed is equal to 0. Now we can multiply both sides of the equation by 4 just to remove the fraction, and so we'll get Y 6 minus 24, multiplied by y cubed is equal to 0. And now we need to solve this for Y. We can factor out a Y cubed out of both of these two terms, and so we'll get Y cubed, multiplied by the quantity of Y cubed. Minus 24 in 20, and that's going to be equal to 0. So to solve for Y, we're going to set each one of these factors equal to 0 and solve for Y. And so, for the first factor we'll have Y cubed is equal to 0, so that means that Y is going to be equal to 0. For the second factor, we'll have Y cubed minus 24 is going to be equal to 0. And solving this for Y, we'll add 24 to both sides, we'll have Y cubed is equal to 24, and then taking the cube root, we'll have Y. is equal to 2 multiplied by the cube root of 3. So now we need to take these Y values and find their corresponding X values. When Y is equal to 0, X is going to be equal to 02 divided by 2, which is going to be equal to 0. And when Y is equal to 2 multiplied by the cube root of 3, our X value is going to be equal to the quantity of 2 multiplied by the cube root of 3 in quantity squared divided by 2, and this is going to be equal to 2 multiplied by the cube root of 9. So, we actually have two points, where we have vertical tangent lines. We have the origin, which is 0.0. And the point of 2 multiplied by the cube root of 9. 2 multiplied by the cube root of 3. Now the question said it wanted the point other than the origin, so we're going to choose our second point as the answer to our problem. And when we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

The folium of Descartes (See Figure 3.27)

c. Find the coordinates of the point A in Figure 3.29 where the folium has a vertical tangent line.

Key Concepts

Vertical Tangent Line

Implicit Differentiation

Folium of Descartes

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