{Use of Tech} Special curves The following classical curves ha...

Question

{Use of Tech} Special curves The following classical curves have been studied by generations of mathematicians. Use analytical methods (including implicit differentiation) and a graphing utility to graph the curves. Include as much detail as possible.

x³ + y³ = 3xy (Folium of Descartes)

x³ + y³ = 3xy (Folium of Descartes)

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Graph the given classical curve using analytical methods. X 3 plus 3 is equal to 6 XY. Dy by DX is equal to 6Y minus 3 x 2, all divided by 32 minus 6 X. Awesome. So it appears for this particular problem, we're given two different pieces of information we're given our function, which is X 3 plus Y 3 is equal to 6 XY. And we're given the derivative of our function it appears, and we're asked to take this information and we're asked to use analytical methods to help graph this classical curve. So now that we know that we're ultimately trying to create a graph for this specific function. Our first step that we need to do in order to solve this problem is we need to identify the curve, and as we should note, this equation represents a folium, a discaris, which is a classic curve with a distinctive loop and has an asymptote. And our second step is we need to find the X intercepts, so we need to set Y equal to 0. So that will mean we can go ahead and write that X to the power of 3. So if we take X to the power of 3 is equal to 0, that will mean that X is equal to 0. Once again, this is when we set. Y is equal to 0, so we plug in a value of 0 in for Y. So thus the only X intercept is going to be at X equals 0. So now, that was our second step. So now for our 3rd step is we need to find the Y intercept now. So to find the Y intercept, we now need to set X equal to 0. So Y to the power 3 equals 0. So that will mean when we rearrange our equation to isolate and solve for Y, Y will be equal to 0. So therefore, the only Y intercept is going to be Y equals 0. Now our fourth step is we need to determine these symmetry, which I'll just put in S in quotation marks. So as we should recall a note for X-axis symmetry, we need to replace the Y with negative Y, and when we do that, the equation will become X 3 minus Y 3 is equal to -6XY. Which is not the same as the original function, so therefore, there will be no symmetry about the X axis. And for Y axis symmetry, we need to replace the X with negative X, and when we do that, the equation will become negative X 3 plus Y 3 is equal to -6XY, which is not the same as the original. So therefore, as a result, there will be no symmetry about the Y axis. So I'll put red X's by both of our functions because it doesn't have symmetry. So it doesn't have symmetry on the X-axion. It doesn't have symmetry on the Y axis. So now, does it have symmetry about the origin. So in order to determine if it has symmetry about the origin, we need to replace X with negative X and Y with negative Y. And when we do that, the function will become negative X 3 minus Y 3 is equal to 6 XY. Which is not the same as the original function, so therefore, there is no symmetry about the origin. So therefore, this curve has no symmetry, so I'll put no with an exclamation point. So there's no symmetry for this curve. So now for our 5th step, we need to determine the asymptotes, which I'll put a little, we'll put a capital A in quotation marks. So, as we should recall, for vertical asymptotes, there is no division by zero occurring, so that will mean that there's no vertical asymptotes because there's no division by zero, there is no vertical asymptotes. So, let's make a note of that. So, VA, vertical asymptotes equals none. Now we need to determine if there's any horizontal asymptotes, which you'll call HA. So VA so we have VA, which is vertical asymptotes, and there's no vertical asymptotes. Now for horizontal asymptotes, HA. As we should recall, as X and Y become large, the cubic term dominates. So therefore, we need to note that the curve does not approach a horizontal line. So that means that there's no horizontal asymptotes. Fantastic. As for our next asymptote we need to look for, which is an oblique or a diagonal asymptote. And we need to recall a note that the equation is cubic and does not qualify, or I should say that the equation is not cubic, therefore, it doesn't simplify to a linear form or doesn't qualify as a linear form. So therefore, there will be no oblique asymptotes, so no diagonal asymptotes will be present. So, therefore, the curve has no asymptotes. So I'll put a no with an exclamation point. So now our 6th step is we need to determine the critical points, which I'll call CP. So in order to find the critical points, we need to set DY by DX equal to 0, which will therefore imply that 6 Y minus 3 x 2 is equal to 0. So when we simplify, we will find that it's going to be 2 Y is equal to x 2. So, therefore, as a result, the critical points satisfy the equation Y equals, once again, we're rearranging the equation to isolate and solve for Y, and when we do that, we will find that Y is equal to X2 divided by 2. So now, our 7th step. we need to substitute into the original equation, so we need to substitute in our value of Y into our original function. And when we do that, we will find that X 3 plus x 2 divided by 23 is equal to 6 X multiplied by x 2 divided by 2. So when we simplify this expression, We will find that it's gonna be equal to X 3 plus x 6 divided by 8 is equal to 3 x 3. So now we need to move all of our terms to one side. So when we do that, we will get that X 3 plus X 6 divided by 8, minus 3 x 3 is equal to 0. So now, our next step is we're gonna be trying to isolate. And sulfur X all by itself, so we're trying to isolate X. So, in order to do that, we need to factor out X to the power of 3. And when we do that, it will give us two possibilities when we factor out X to the power 3. So we will find that X to the power of 3 is equal to 0, so that will imply that X is equal to 0, or when we factor it out, we'll get X 3 divided by 8, minus 2 is equal to 0, which will mean that X to the power 3 is equal to 16, which will mean that X is equal to the cube root of 16. Which will give us an answer of only around 23 deci places, 2.520. So now, Once again, we factored out our 3, so I did it for, I skipped ahead and did it for the two separate possibilities, but initially when you pull out, here I'll make a note, I'll put a start. When we pull out X to the 3, it'll be X to the power 3 multiplied by X to the 3 divided by 8, minus 2 is equal to 0. Fantastic. So now we need to solve for Y, which is our step 8. So step 8, we're solving for Y. So for each X, we need to solve for Y when Y is equal to X to the power 2 divided by 2. So when X is equal to 0, Y is equal to 0, and when X is equal to the cube root of 16, that will mean that Y is going to be equal to the cube root of 16. To the power to divided by 2, which will mean that Y is going to be equal to 3.175 when you evaluate the value for Y when we round to 3 decimal points. So therefore, without a doubt, the critical points will be 0.0 and 2.520.3.175. Fantastic. So now we can officially sketch the graph, so we need to note that the curve will pass through the origin point, which is 0.0, and will behave non-symmetrically with a changing slope that is defined by Dy by DX. So when we use a graphing utility to plot the curve X 3 plus Y 3 equals 6 xy, we will produce the following graph as a result. So, as you could see, like we were discussing earlier, when we initially discussed the problem, we have a loop de loop. We have a little loop in the center of our curve, and our curve is represented in blue. And then we have our two critical points represented in red, so we have it at 00, which is the origin point. And then we have our second critical point, which we have it written. As 16 to the power of 1/3, 32 of 13, so that's another way of saying the cube root of 16 and the cube root of 32. Fantastic. And that's it. So when we draw a smooth curve to connect our two critical points, we will see that it forms a loop de loop with no asymptotes present. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

{Use of Tech} Special curves The following classical curves have been studied by generations of mathematicians. Use analytical methods (including implicit differentiation) and a graphing utility to graph the curves. Include as much detail as possible.

x³ + y³ = 3xy (Folium of Descartes)

Key Concepts

Implicit Differentiation

Graphing Utility

Classical Curves

Watch next

{Use of Tech} Special curves The following classical curves have been studied by generations of mathematicians. Use analytical methods (including implicit differentiation) and a graphing utility to graph the curves. Include as much detail as possible.x³ + y³ = 3xy (Folium of Descartes)

Key Concepts

Implicit Differentiation

Graphing Utility

Classical Curves

Watch next

{Use of Tech} Special curves The following classical curves have been studied by generations of mathematicians. Use analytical methods (including implicit differentiation) and a graphing utility to graph the curves. Include as much detail as possible.

x³ + y³ = 3xy (Folium of Descartes)