# Implicit differentiation question

1. Oct 8, 2006

### donjt81

I have this question in which I know I probably have to use implicit differentiation but I have no idea how to do this can someone give me a hint to get started. all the implicit differentiation problems I have done only have a combination of x and y but this one has x, y and t.

find dy/dx for x = 4sint, y= 4cost

2. Oct 8, 2006

Those are parametric equations. So $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

3. Oct 8, 2006

### donjt81

Ohh ok soo... this is what I did

d/dt x = d/dt 4sint
dx/dt = 4cost

d/dt y = d/dt 4cost
dy/dt = -4sint

dy/dx = -4sint/4cost
= -tant

does that look right?

4. Oct 8, 2006

yep, that is correct.

5. Oct 8, 2006

### donjt81

thanks a lot!!!

6. Oct 8, 2006

### donjt81

I have another similar question... i was wondering if you could help me out

so now i have a problem in "related rates" and I have to use the volume of a cylinder formula.

v = pi * r^2 * h

rate of change of the radius (dr/dt) is given and the rate of change of the height (dh/dt) is given. I am supposed to find the rate of change of the volume (dv/dt). so i thought finding the d/dt of each side of the above equation will get me the answer but i dont know how to differentiate the right side.... this is what i am looking at

d/dt v = d/dt (pi * r^2 * h)

can you help me?

7. Oct 8, 2006

So $$V = \pi r^{2}h$$. Then $$\frac{dV}{dt} = 2\pi r \frac{dh}{dt}$$. You have to eliminate one of the variables by using similar triangles. So $$\frac{r}{h} = \frac{x}{y}$$.

Last edited: Oct 8, 2006
8. Oct 8, 2006

### donjt81

but then i wont be able to use dr/dt. Why would they give me dr/dt in the problem.

9. Oct 8, 2006

Whoops, it should be $$\frac{dV}{dt} = \pi r^{2} \frac{dh}{dt} + 2r\pi h \frac{dr}{dt}$$.

Last edited: Oct 8, 2006
10. Oct 8, 2006

### donjt81

ok that makes sense... i just wanted to be sure i did it right because the answer is a big number. this is the question...

the question is:
The radius of a right circular cylinder is increasing at the rate of 6in/sec, while the height is decreasing at the rate of 8in/sec. at what rate is the volume of the cylinder changing when the radius is 13in and the height is 12 in.

so i used your help and plugged in the values

dv/dt = (3.14 * 13^2 * -8) + (2 * 13 * 3.14 * 12 * 6)
= 1633.6 in^3/sec

its kind of a big number but i guess it is normal for volume to increase at that rate.. i dont know... what do you think.

11. Oct 8, 2006