# Differential Calculus Question

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1. Jul 31, 2015

### hiineko

If z = e ^ (xy ^ 2), x = tcost, and y = tsint compute dx / dt for t = pi / 2

I kind of lost in this difficult question pls help

I tried putting down the xy but using ln

lnz = xy^2

Product rule? Or what. This is my first time encountering this kind of question

Last edited: Jul 31, 2015
2. Jul 31, 2015

### PWiz

Are you sure that you need to calculate $\frac{dx}{dt}$? If so, then the first and 3rd equations are irrelevant - just calculate $\frac{d}{dt} (t cost)$ using the product rule, and plug in $\frac π {2}$ for $t$ into the derivative.

3. Jul 31, 2015

### hiineko

Hmm that make sense
but in my hw it is dx/dt idk if that is a correction

I also think it is a dz/dt

4. Jul 31, 2015

### PWiz

For dz/dt, why don't you substitute the values of x and y in terms of t for the expression of z first?

5. Jul 31, 2015

### hiineko

hmmmm I tried this

Product rule?

6. Jul 31, 2015

### HallsofIvy

Staff Emeritus
If the question is just "what is dx/dt", then the answer is just d(t cos(t))/dt= cos(t)- t sin(t), by the product rule.

But perhaps the problem was to find dz/dt and you are asked to find dx/dt as part of that.

By the "chain rule", $dz/dt= (\partial z/\partial x)(dx/dt)+ (\partial z/\partial y)(dy/dt)$

Since $z= e^{xy^2}$, $\partial z/\partial x= y^2e^{xy^2}$ and $\partial z/\partial y= 2xy e^{xy^2}$

As before, $dx/dt= cos(t)- t sin(t)$ and, since $y= t sin(t)$, $dy/dt= sin(t)+ t cos(t)$.

7. Jul 31, 2015

### PWiz

There are two ways to proceed from here. Are you familiar with logarithmic differentiation? If yes, then just differentiate the left side with respect to t using chain rule, and the right side in the regular product rule manner.

If not, then convert this back into exponential form and use chain rule in combination with the product rule by letting the exponent equal to a new intermediate variable, such as u.

8. Jul 31, 2015

### hiineko

I think I can do the Logarithmic Differentiation

d/dx(lnu) = 1/u * du/dx ??

9. Jul 31, 2015

### PWiz

This is true for the LHS, except that instead of u there will be a z, and instead of x we will have a t - there is no need to introduce a new variable for differentiating the LHS.

10. Jul 31, 2015

### hiineko

Hmm can you lead me there? I think I can understand if I can grasp more of it thank you!

11. Jul 31, 2015

### PWiz

You already did that. The equation becomes $$\frac 1 {z} \frac{dz}{dt} = \frac{d}{dt} (\frac{t^4 (sin 2t)^2}{4})$$ , and you can proceed from here by applying the product rule.

Last edited: Jul 31, 2015
12. Jul 31, 2015

### hiineko

I'll chain rule the RHS?
I'll took the constant out first so
1/4(t^4(sin2t)^2)?
I don't know if that's what you meant

13. Jul 31, 2015

### PWiz

Oh, my bad. I meant use the product rule.

14. Jul 31, 2015

### hiineko

Soooooo

I'll just product rule this then substitue pi/2?

15. Jul 31, 2015

### PWiz

Yes.

16. Jul 31, 2015

### hiineko

I got 0.3448 is it right?

17. Jul 31, 2015

### Qwertywerty

Are you sure that works out to give 0.3448 ?

18. Aug 1, 2015

### PWiz

So did you get an answer?

19. Aug 1, 2015

### hiineko

I got 0.3448 I don't know if the answer is correct but at least I understand the concept which is more important so thank you mates!

20. Aug 1, 2015

### PWiz

EDIT: The answer is close, but not 0.3448 . Can show the derivative that you got?

Last edited: Aug 1, 2015