Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Implicit differentiation question

  1. Oct 8, 2006 #1
    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

    thanks in advance...
  2. jcsd
  3. Oct 8, 2006 #2
    Those are parametric equations. So [tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]
  4. Oct 8, 2006 #3
    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?
  5. Oct 8, 2006 #4
    yep, that is correct.
  6. Oct 8, 2006 #5
    thanks a lot!!!
  7. Oct 8, 2006 #6
    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?

    Thanks in advance
  8. Oct 8, 2006 #7
    So [tex] V = \pi r^{2}h [/tex]. Then [tex] \frac{dV}{dt} = 2\pi r \frac{dh}{dt} [/tex]. You have to eliminate one of the variables by using similar triangles. So [tex] \frac{r}{h} = \frac{x}{y} [/tex].
    Last edited: Oct 8, 2006
  9. Oct 8, 2006 #8
    but then i wont be able to use dr/dt. Why would they give me dr/dt in the problem.
  10. Oct 8, 2006 #9
    Whoops, it should be [tex] \frac{dV}{dt} = \pi r^{2} \frac{dh}{dt} + 2r\pi h \frac{dr}{dt} [/tex].
    Last edited: Oct 8, 2006
  11. Oct 8, 2006 #10
    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.
  12. Oct 8, 2006 #11
    its correct
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook