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Synne's question at Yahoo! Answers regarding the slope of an astroid

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MarkFL

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Feb 24, 2012
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Here is the question:

Calculus- find the points on the astroid where the tangent line has slope 1?

The equation of the astroid is x^(2/3) + y^(2/3) =1

I found that the derivative is -(y/x)^(1/3) = y'

I'm not sure how to get the points where the tangent line has slope one though- I tried plugging in one for y'; I just got -x =y, which seems right considering the answers, but I'm not sure how to get specific points.

According to the book, the answers are (2^(1/2)/4, 2^(1/2)/-4) and (2^(1/2)/-4,2^(1/2)/4)

I'd appreciate any help in the right direction.
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Synne,

We are given the curve:

\(\displaystyle x^{\frac{2}{3}}+y^{\frac{2}{3}}=1\)

Implicitly differentiating with respect to $x$, we find:

\(\displaystyle \frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0\)

Multiply through by \(\displaystyle \frac{3}{2}\):

\(\displaystyle x^{-\frac{1}{3}}+y^{-\frac{1}{3}}\frac{dy}{dx}=0\)

Subtract through by \(\displaystyle x^{-\frac{1}{3}}\):

\(\displaystyle y^{-\frac{1}{3}}\frac{dy}{dx}=-x^{-\frac{1}{3}}\)

Multiply through by \(\displaystyle y^{\frac{1}{3}}\):

\(\displaystyle \frac{dy}{dx}=-\left(\frac{y}{x} \right)^{\frac{1}{3}}\)

Equate this result to $1$:

\(\displaystyle -\left(\frac{y}{x} \right)^{\frac{1}{3}}=1\)

Cube both sides:

\(\displaystyle -\frac{y}{x}=1\)

Multiply through by $-x$:

\(\displaystyle y=-x\)

Now, substitute for $y$ into the astroid:

\(\displaystyle x^{\frac{2}{3}}+(-x)^{\frac{2}{3}}=1\)

Now, if we observe that:

\(\displaystyle (-x)^{\frac{2}{3}}=\left((-x)^2 \right)^{\frac{1}{3}}=\left(x^2 \right)^{\frac{1}{3}}=x^{\frac{2}{3}}\)

we now have:

\(\displaystyle 2x^{\frac{2}{3}}=1\)

Cube both sides:

\(\displaystyle 8x^2=1\)

\(\displaystyle x^2=\frac{1}{8}\)

\(\displaystyle x=\pm\frac{1}{2\sqrt{2}}\)

Now since $y=-x$, our two points are then:

\(\displaystyle (x,y)=\left(\pm\frac{1}{2\sqrt{2}}, \mp\frac{1}{2\sqrt{2}} \right)\)
 

Viridian

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Feb 27, 2014
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Thank you, that was very helpful.

Unfortunately for me, on my Calc test the other day, my teacher gave us a question similar to this one, except the original equation could not be rearranged to isolate y. If I remember correctly, the question was (x2 + y2)2/3 = x2 - y2.

Would you have any suggestions of any techniques to try to solve a question like this without plugging in the value of y from the original equation?
 
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MarkFL

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Do you happen to recall what the value of the slope was?
 

Viridian

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Feb 27, 2014
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Oh yes, it was where the slope was zero.
 
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MarkFL

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Oh yes, it was where the slope was zero.
I was hoping you were going to say zero, because that makes things considerably simpler.

Have you computed \(\displaystyle \frac{dy}{dx}\) ?
 

Viridian

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Feb 27, 2014
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I can't recall what I wrote on the test exactly, of course, but here's what I did just now:
(x2 + y2)2/3 = x2 - y2
2/3(x2 + y2)-1/3)(2x + 2yy') = 2x - 2yy'
2(2x+2yy')= (2x-2yy')(3((x2 + y2)-1/3)
2(2x+2yy')/(2x-2yy')=3((x2 + y2)-1/3
[2(2x+2yy')/3(2x-2yy')]-3/2=r

I tried some more things on paper but I couldn't figure out how to isolate y'.
 
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MarkFL

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Okay, this is what I did. We are given:

\(\displaystyle \left(x^2+y^2 \right)^{\frac{2}{3}}=x^2-y^2\)

Implicitly differentiate with respect to $x$:

\(\displaystyle \frac{2}{3}\left(x^2+y^2 \right)^{-\frac{1}{3}}(2x+2yy')=2x-2yy'\)

This agrees with your first step, so I know you are correctly applying the power and chain rules. However, your next algebraic step is incorrect. You essentially tried to use:

\(\displaystyle ab^c=d\implies a=db^c\)

And this is not true in general. For my next steps, I multiplied through by \(\displaystyle \frac{3}{2}\):

\(\displaystyle \left(x^2+y^2 \right)^{-\frac{1}{3}}(2x+2yy')=3x-3yy'\)

Distribute on the left side:

\(\displaystyle 2x\left(x^2+y^2 \right)^{-\frac{1}{3}}+2yy'\left(x^2+y^2 \right)^{-\frac{1}{3}}=3x-3yy'\)

Now, what you want to do to continue is to arrange the equation such that all the terms having $y'$ as a factor are on one side, and everything else is on the other. Then factor out $y'$ and divide by the other factor to isolate $y'$. Can you proceed?
 

Viridian

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Feb 27, 2014
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Okay, this is what I did. We are given:

\(\displaystyle \left(x^2+y^2 \right)^{\frac{2}{3}}=x^2-y^2\)

Implicitly differentiate with respect to $x$:

\(\displaystyle \frac{2}{3}\left(x^2+y^2 \right)^{-\frac{1}{3}}(2x+2yy')=2x-2yy'\)

This agrees with your first step, so I know you are correctly applying the power and chain rules. However, your next algebraic step is incorrect. You essentially tried to use:

\(\displaystyle ab^c=d\implies a=db^c\)

And this is not true in general. For my next steps, I multiplied through by \(\displaystyle \frac{3}{2}\):

\(\displaystyle \left(x^2+y^2 \right)^{-\frac{1}{3}}(2x+2yy')=3x-3yy'\)

Distribute on the left side:

\(\displaystyle 2x\left(x^2+y^2 \right)^{-\frac{1}{3}}+2yy'\left(x^2+y^2 \right)^{-\frac{1}{3}}=3x-3yy'\)

Now, what you want to do to continue is to arrange the equation such that all the terms having $y'$ as a factor are on one side, and everything else is on the other. Then factor out $y'$ and divide by the other factor to isolate $y'$. Can you proceed?
Oh, yes. That makes sense.
So

\(\displaystyle 2yy'\left(x^2+y^2 \right)^{-\frac{1}{3}}+3yy'=3x - 2x\left(x^2+y^2 \right)^{-\frac{1}{3}}\)

\(\displaystyle y'(2y\left(x^2+y^2 \right)^{-\frac{1}{3}}+3y) = 3x - 2x\left(x^2+y^2 \right)^{-\frac{1}{3}}\)

\(\displaystyle y'= {\frac{x(3\left(x^2+y^2 \right)^{\frac{1}{3}}-2)}{y(3\left(x^2+y^2 \right)^{\frac{1}{3}}+2)}}\)

This is at the point where y' would be set to 0 to represent the slope, yes? Of course, there's still y and x values for both the denominator and numerator unfortunately. I didn't do it wrong, did I?
 

MarkFL

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Feb 24, 2012
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Good job! And I don't think I've ever seen someone pick up the use of $\LaTeX$ so quickly either. (Muscle)

So, if this is equated to zero, what do it imply? For example, if:

\(\displaystyle \frac{a}{b}=0\)

What may we conclude regarding $a$ and $b$?
 

Viridian

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Feb 27, 2014
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Good job! And I don't think I've ever seen someone pick up the use of $\LaTeX$ so quickly either. (Muscle)

So, if this is equated to zero, what do it imply? For example, if:

\(\displaystyle \frac{a}{b}=0\)

What may we conclude regarding $a$ and $b$?
Thanks, it's because I'm used to html. :p

Well, since b cannot equal 0, then a must be the one to equal 0, correct? So are you implying that:
\(\displaystyle x=0\)
and
\(\displaystyle 3(x^2+y^2)^{\frac {1}{3}}-2 =0\)

I remember in a previous question like this x = 0 was not a real answer; is it the case here as well?
 

MarkFL

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Feb 24, 2012
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As long as the denominator has no real roots, then we may equate the numerator to zero with no worries that we might have an indeterminate form on the left.
 

Viridian

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Feb 27, 2014
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As long as the denominator has no real roots, then we may equate the numerator to zero with no worries that we might have an indeterminate form on the left.
So then, to solve for x in the second equation, what would we do? I'm not sure how to cancel out either the y2 or x2 in order to solve for one of them, as the original equation cannot be isolated for y.
 

MarkFL

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Equating the numerator to zero, we have:

\(\displaystyle x(3\left(x^2+y^2 \right)^{\frac{1}{3}}-2)=0\)

And so we immediately see that we have $x=0$. What do you find when you substitute this value for $x$ into the original equation?

And we also must consider:

\(\displaystyle 3\left(x^2+y^2 \right)^{\frac{1}{3}}-2=0\)

Since the original equation has only $x^2$ and $y^2$ as variable terms in it, you only need to solve for either. What do you find?
 

Viridian

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Feb 27, 2014
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Equating the numerator to zero, we have:

\(\displaystyle x(3\left(x^2+y^2 \right)^{\frac{1}{3}}-2)=0\)

And so we immediately see that we have $x=0$. What do you find when you substitute this value for $x$ into the original equation?

And we also must consider:

\(\displaystyle 3\left(x^2+y^2 \right)^{\frac{1}{3}}-2=0\)

Since the original equation has only $x^2$ and $y^2$ as variable terms in it, you only need to solve for either. What do you find?
Okay, so if we plug in x = 0 into \(\displaystyle 3\left(x^2+y^2 \right)^{\frac{1}{3}}-2=0\)

\(\displaystyle 3\left(y^2 \right)^{\frac{1}{3}}-2=0\)
\(\displaystyle y^{\frac{2}{3}}=2/3\)
\(\displaystyle y^2=(2/3)^3\)
\(\displaystyle y=+/-(8/27)^{\frac{1}{2}}\)

If we substitute that into the original equation:

\(\displaystyle (x^2+y^2)^{\frac{2}{3}}=x^2-y^2\)
\(\displaystyle (x^2+(2/3)^3)^{\frac{2}{3}}=x^2-(2/3)^3\)
\(\displaystyle (x^2+(2/3)^3)^2=(x^2-(2/3)^3))^3\)
...
I then tried to expand it out on paper. It did not turn out well.

Okay well then if x = 0 for the original equation:
\(\displaystyle (y^2)^{\frac{2}{3}}=-y^2\)
\(\displaystyle y(y^{\frac{1}{3}})=-y^2\)
\(\displaystyle y^{\frac{1}{3}}=-y\)
\(\displaystyle 0=-y^3-y\)
\(\displaystyle 0= y(-y^2-1)\)

y = 0, \(\displaystyle -y^2 = 1\)
If \(\displaystyle -y^2=1\), then:
\(\displaystyle -y = 1^{\frac{1}{2}}\)
\(\displaystyle y= 1, y = -1\)

Plugging that into the original equation brought the same problems as with plugging in those other answers... I must be doing something wrong, especially as this is a no-calculator test.
 

MarkFL

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Feb 24, 2012
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You want to substitute into the original equation the values you find for either variable from the implication of equating the derivative to zero.

The first implication is $x=0$. Plugging that into our original equation, we obtain:

\(\displaystyle y^{\frac{4}{3}}=-y^2\)

So, what must $y$ be? Just use the fact that for all real $y$ we must have:

\(\displaystyle 0\le y^{\frac{4}{3}}\) and \(\displaystyle 0\ge -y^2\)

Does the value you find for $y$ cause a problem with the expression for $y'$?

Now, for the second implication, which comes from:

\(\displaystyle 3\left(x^2+y^2 \right)^{\frac{1}{3}}-2=0\)

We find:

\(\displaystyle x^2+y^2=\left(\frac{2}{3} \right)^3\)

which can also be arranged as:

\(\displaystyle y^2=\left(\frac{2}{3} \right)^3-x^2\)

Using the first form on the left side of the original equation, and the second form on the right side, we find:

\(\displaystyle \left(\frac{2}{3} \right)^2=x^2-\left(\left(\frac{2}{3} \right)^3-x^2 \right)\)

\(\displaystyle 2x^2=\left(\frac{2}{3} \right)^2+\left(\frac{2}{3} \right)^3\)

\(\displaystyle x^2=\frac{10}{27}\)

But...this implies:

\(\displaystyle y^2=\left(\frac{2}{3} \right)^3-\frac{10}{27}=-\frac{2}{27}\)

And there is no real $y$ that satisfies this equation.

So, what can you conclude from all of this?
 

Viridian

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Feb 27, 2014
14
You want to substitute into the original equation the values you find for either variable from the implication of equating the derivative to zero.

The first implication is $x=0$. Plugging that into our original equation, we obtain:

\(\displaystyle y^{\frac{4}{3}}=-y^2\)

So, what must $y$ be? Just use the fact that for all real $y$ we must have:

\(\displaystyle 0\le y^{\frac{4}{3}}\) and \(\displaystyle 0\ge -y^2\)

Does the value you find for $y$ cause a problem with the expression for $y'$?

Now, for the second implication, which comes from:

\(\displaystyle 3\left(x^2+y^2 \right)^{\frac{1}{3}}-2=0\)

We find:

\(\displaystyle x^2+y^2=\left(\frac{2}{3} \right)^3\)

which can also be arranged as:

\(\displaystyle y^2=\left(\frac{2}{3} \right)^3-x^2\)

Using the first form on the left side of the original equation, and the second form on the right side, we find:

\(\displaystyle \left(\frac{2}{3} \right)^2=x^2-\left(\left(\frac{2}{3} \right)^3-x^2 \right)\)

\(\displaystyle 2x^2=\left(\frac{2}{3} \right)^2+\left(\frac{2}{3} \right)^3\)

\(\displaystyle x^2=\frac{10}{27}\)

But...this implies:

\(\displaystyle y^2=\left(\frac{2}{3} \right)^3-\frac{10}{27}=-\frac{2}{27}\)

And there is no real $y$ that satisfies this equation.

So, what can you conclude from all of this?
I can conclude that my teacher is crazy trying to make a bunch of grade elevens solve this on a marked test without explaining it at all.

Sorry. Jokes aside:

Are you implying that there are no real solutions for this question then?
 

MarkFL

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Feb 24, 2012
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Well, what I conclude is that while the point $(0,0)$ is on the given curve, the slope is indeterminate since we get 0/0. So add that to the other implication, and I would say there is no point on the given curve for which the slope of the tangent line is zero.
 

Viridian

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Feb 27, 2014
14
Well, what I conclude is that while the point $(0,0)$ is on the given curve, the slope is indeterminate since we get 0/0. So add that to the other implication, and I would say there is no point on the given curve for which the slope of the tangent line is zero.

So I graphed it on a graphing calculator, and lo and behold, a graph where there is no horizontal tangent line.

At this point, I'm hoping I just remembered it wrong; but it's probably just my teacher being a horrible troll again. An eight mark question with no real answer? Sounds just up his alley.

Thank you very much for your help.
 

MarkFL

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Feb 24, 2012
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Actually, I would have to say this is a great question, as it makes you think about the implications of your results. (Tmi)(Bug)(Headbang)(Drunk)(Yes)
 

Viridian

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Feb 27, 2014
14
Actually, I would have to say this is a great question, as it makes you think about the implications of your results. (Tmi)(Bug)(Headbang)(Drunk)(Yes)
Great as a question in general, no~t so great for a question worth a third of the marks on a test. (Headbang) (Dull)