Mixing
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Find coordinate in first quadrant which tangent line to $x^3-xy+y^3=0$ has slope 0
Find coordinate in first quadrant which tangent line to $x^3-xy+y^3=0$ has slope 0
First, I do implicit differentiation:
$frac{3x^2-y}{x-3y^2}=y'$
so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.
But then I graph it on a software and i get the following image-
Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right
calculus derivatives implicit-differentiation
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
asked Nov 20 '18 at 21:53
John Rawls
1,253519
add a comment |
Find coordinate in first quadrant which tangent line to $x^3-xy+y^3=0$ has slope 0
First, I do implicit differentiation:
$frac{3x^2-y}{x-3y^2}=y'$
so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.
But then I graph it on a software and i get the following image-
Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right
calculus derivatives implicit-differentiation
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
asked Nov 20 '18 at 21:53
John Rawls
1,253519
add a comment |
Find coordinate in first quadrant which tangent line to $x^3-xy+y^3=0$ has slope 0
First, I do implicit differentiation:
$frac{3x^2-y}{x-3y^2}=y'$
so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.
But then I graph it on a software and i get the following image-
Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right
calculus derivatives implicit-differentiation
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
asked Nov 20 '18 at 21:53
John Rawls
1,253519
Find coordinate in first quadrant which tangent line to $x^3-xy+y^3=0$ has slope 0
First, I do implicit differentiation:
$frac{3x^2-y}{x-3y^2}=y'$
so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.
But then I graph it on a software and i get the following image-
Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right
calculus derivatives implicit-differentiation
calculus derivatives implicit-differentiation
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
asked Nov 20 '18 at 21:53
John Rawls
1,253519
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
asked Nov 20 '18 at 21:53
John Rawls
1,253519
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
edited Nov 21 '18 at 14:14
André 3000
12.5k22042
12.5k22042
asked Nov 20 '18 at 21:53
John Rawls
1,253519
asked Nov 20 '18 at 21:53
John Rawls
1,253519
asked Nov 20 '18 at 21:53
John Rawls
1,253519
1,253519
add a comment |
add a comment |
2 Answers
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We have that
$$x^3-xy+y^3=0implies 3x^2dx-dx-xdy+3y^2dy=0 implies frac{dy}{dx}=frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 iff t^3-3t^3+27t^6=0iff t^3(27t^3-2)=0$$
that is
$$(x,y)=left(frac{sqrt[3]2}{3},sqrt[3]4right)$$
answered Nov 20 '18 at 21:56
gimusi
1
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
add a comment |
You solved only half of the problem. You have that the derivative is $0$, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.
Just plug in $y=3x^2$ into your original equation, and solve for $x$
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
We have that
$$x^3-xy+y^3=0implies 3x^2dx-dx-xdy+3y^2dy=0 implies frac{dy}{dx}=frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 iff t^3-3t^3+27t^6=0iff t^3(27t^3-2)=0$$
that is
$$(x,y)=left(frac{sqrt[3]2}{3},sqrt[3]4right)$$
answered Nov 20 '18 at 21:56
gimusi
1
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
add a comment |
We have that
$$x^3-xy+y^3=0implies 3x^2dx-dx-xdy+3y^2dy=0 implies frac{dy}{dx}=frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 iff t^3-3t^3+27t^6=0iff t^3(27t^3-2)=0$$
that is
$$(x,y)=left(frac{sqrt[3]2}{3},sqrt[3]4right)$$
answered Nov 20 '18 at 21:56
gimusi
1
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
add a comment |
We have that
$$x^3-xy+y^3=0implies 3x^2dx-dx-xdy+3y^2dy=0 implies frac{dy}{dx}=frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 iff t^3-3t^3+27t^6=0iff t^3(27t^3-2)=0$$
that is
$$(x,y)=left(frac{sqrt[3]2}{3},sqrt[3]4right)$$
answered Nov 20 '18 at 21:56
gimusi
1
We have that
$$x^3-xy+y^3=0implies 3x^2dx-dx-xdy+3y^2dy=0 implies frac{dy}{dx}=frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 iff t^3-3t^3+27t^6=0iff t^3(27t^3-2)=0$$
that is
$$(x,y)=left(frac{sqrt[3]2}{3},sqrt[3]4right)$$
answered Nov 20 '18 at 21:56
gimusi
1
edited Nov 20 '18 at 21:59
answered Nov 20 '18 at 21:56
gimusi
1
answered Nov 20 '18 at 21:56
gimusi
1
answered Nov 20 '18 at 21:56
gimusi
1
1
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
add a comment |
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
sorry i dont understand what the T is? because this isnt a parametric question right?
– John Rawls
Nov 20 '18 at 21:58
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
@JohnRawls The condition $y=3x^2$ corresponds to a parabola, we need to find the intersection with the curve to find the point.
– gimusi
Nov 20 '18 at 22:00
add a comment |
You solved only half of the problem. You have that the derivative is $0$, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.
Just plug in $y=3x^2$ into your original equation, and solve for $x$
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
add a comment |
You solved only half of the problem. You have that the derivative is $0$, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.
Just plug in $y=3x^2$ into your original equation, and solve for $x$
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
add a comment |
You solved only half of the problem. You have that the derivative is $0$, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.
Just plug in $y=3x^2$ into your original equation, and solve for $x$
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
You solved only half of the problem. You have that the derivative is $0$, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.
Just plug in $y=3x^2$ into your original equation, and solve for $x$
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
answered Nov 20 '18 at 22:00
Andrei
11.3k21026
11.3k21026
add a comment |
add a comment |
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