Differentiation of (x^3)(y^2)=128?

[zfunk]

Differentiation of (y^4)(x^2)=128?

Differentiation of (y^4)(x^2)=128, and then given dy/dx = 4 show c = -2d and find the value of c and d.
been stuck on this question for a while now, any help would be appreciated thanks!

 

[MarkFL]

Hello and welcome to MHB, zfunk!

I have moved your topic to the Calculus sub-forum, as it involves differentiation.

Can you show what you have tried so far? You want to implicitly differentiate with respect to $x$. This means you are going to need to apply the product, power and chain rules. Can you find \(\displaystyle \frac{dy}{dx}\)?

 

[zfunk]

thanks! when i differentiated i got : (-3x^2)(y^2)/(2x^3)(y).
i also forgot to mention in the question that the point where dy/dx=3 is (a,b).
I feel im right so far but when it comes to working out the coordinates, i cant get an answer, thanks again!

 

[MarkFL]

Yes, you have differentiated correctly.

You now want to simplify by canceling or dividing out factors common to the numerator and denominator.

I appreciate the clarification, because initially I did not know where $a$ and $b$ came from, until I found \(\displaystyle \frac{dy}{dx}\).

 

[zfunk]

i managed to cancel down to give 6a^3/-3a^2=b which then simplified to b=-2a,
however when i sub the b=-2a into the question i dont get an answer? does that mean a=0? or am i doing it all wrong? thanks

 

[MarkFL]

This is the way I would work the problem:

We are given:

\(\displaystyle x^3y^2=128\)

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

\(\displaystyle x^32y\frac{dy}{dx}+3x^2y^2=0\)

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

Now, we are told to equate this to 3 at the point $(a,b)$:

\(\displaystyle \left.\frac{dy}{dx}\right|_{(x,y)=(a,b)}=3\)

\(\displaystyle -\frac{3b}{2a}=3\)

Can you proceed?

 

[zfunk]

i get this far but when subbing in b=-2a i dont get a proper answer and cant understand what going wrong

 

[MarkFL]

Okay, we've got $x=a$ and $y=b=-2a$, now subbing those in:

\(\displaystyle \frac{dy}{dx}=-\frac{3y}{2x}=-\frac{3(-2a)}{2a}=3\)

and this checks with what we were given.

 

[zfunk]

however the question asks for a value for a and b. so what would they be?

 

[MarkFL]

Originally we were given:

\(\displaystyle x^3y^2=128\)

and we found:

\(\displaystyle y=-2x\)

and so substituting for $y$, what do you find?

 

[zfunk]

i think i get a=0? but not 100% on that

 

[MarkFL]

Can you do the substitution? You should find a non-zero value for $a$. I want to see your work when you make the substitution, to see how you are getting zero for $a$.

 

[zfunk]

i get 6x/2x=3 so 3=3, so does that mean a =1? ahh this is soo confusing!

 

[MarkFL]

You have:

\(\displaystyle x^3y^2=128\)

\(\displaystyle y=-2x\)

and so:

\(\displaystyle x^3(-2x)^2=128\)

Now solve for $x$.

 

[zfunk]

i got x=2! thanks! however in the question it says the point P lies on curve V and has coordinates (a,b). So i was trying to sub the b=-2a into the dy/dx equation

 

[MarkFL]

You obtained \(\displaystyle b=-2a\) from the equation:

\(\displaystyle \frac{dy}{dx}=-\frac{3y}{2x}\)

so substituting this result back into it will only result in an identity as I showed above.

You now have that \(\displaystyle a=2\), so what is $b$?

By the way, you will get much better help if you give the problems exactly as stated in their entirety up front. Giving the entire problem, along with the directions, and all work you have done is the best way to get prompt, pertinent help.

I'm not "fussing," but rather offering you helpful advice.

 

[zfunk]

i got b=-4! thanks wil make sure i do! really appreciate you taking time out to help me here thanks!!

 

[MarkFL]

Yes, the point $(2,-4)$ is the only point on the implicitly defined curve at which the slope of the tangent line is $3$.