# Thread: [SOLVED] Implicit diff question.

1. ## [SOLVED] Implicit diff question.

I'm struggling with this question - I think I'm ok with the calculus its just the maniplutaion. Anyway, your help and guidence would be very much appreciated.

The equ. of a curve is (6y^2) = (e^x)(y^3) - 2e^4x.

Given that (a,b) is the point on the curve at which dy/dx = 0,
show that b = 2e^a

By substituting into the equ. obtain a further relationship between a and b, and hence find the values of a and b.

OK

So by implicit diff. and rearranging

(e^x(y^3 - (8e^3x)) / 3y(4 - (e^x)y)) = 0

I'm not really sure how to show the rest though..

can anyone offer some help?

D

2. Originally Posted by dojo
I'm struggling with this question - I think I'm ok with the calculus its just the maniplutaion. Anyway, your help and guidence would be very much appreciated.

The equ. of a curve is (6y^2) = (e^x)(y^3) - 2e^4x.

Given that (a,b) is the point on the curve at which dy/dx = 0,
show that b = 2e^a

By substituting into the equ. obtain a further relationship between a and b, and hence find the values of a and b.

OK

So by implicit diff. and rearranging

(e^x(y^3 - (8e^3x)) / 3y(4 - (e^x)y)) = 0
It looks like you solved for y' and then set it equal to 0. It would have been much easier to just put y'= 0 before solving for it.

But, in any case, you should know that, in order that a fraction be equal to 0, the numerator must be 0- that equation leads immediately to [tex](e^x(y^3- 8e^{3x})= 0[tex] and, since $e^x$ is never 0, $y^3= 8e^{3x}$.

Since this is at (a, b), let x= a, y= b and solve for b.

I'm not really sure how to show the rest though..

can anyone offer some help?

D

3. Hi dojo,

you differentiated nicely,

good suggestion from HallsofIvy,
notationally quite compact...

$6y^2=e^xy^3-2e^{4x}$

$6y'2y=e^xy^3+e^xy'3y^2-8e^{4x}$

set y'=0

$0=e^xy^3-8e^{4x}\ \Rightarrow\ e^x\left(y^3-8e^{3x}\right)=0\ \Rightarrow\ y^3=8e^{3x}=2^3\left(e^x\right)^3=\left(2e^x\right )^3$

$\Rightarrow\ y=2e^x$

4. Thanks for the tips on diff - I see its much easier that way!

Regarding the subbing in of y = 2e^x into the original equ.
I get 24e^(2x) = 6e^(4x)

i cant see a relationship between them at all?

5. Originally Posted by dojo
Thanks for the tips on diff - I see its much easier that way!

Regarding the subbing in of y = 2e^x into the original equ.
I get 24e^(2x) = 6e^(4x)

i cant see a relationship between them at all?
$24e^{2x} - 6e^{4x} = 0$

$6e^{2x}(4 - e^{2x}) = 0$

finish

6. Originally Posted by dojo
Thanks for the tips on diff - I see its much easier that way!

Regarding the subbing in of y = 2e^x into the original equ.
I get 24e^(2x) = 6e^(4x)

i cant see a relationship between them at all?
Just a few more lines and you have it, dojo,

$24e^{2x}=6e^{4x}$

$\frac{24}{6}=\frac{e^{4x}}{e^{2x}}={e^{4x-2x}}=e^{2x}$

$4=e^{2x}=\left(e^x\right)^2$

$e^x=2$

$x=ln2$

$y=2e^{ln2}=4$