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  1. J

    Trig Identities:

    Look at your first "obvious trig identity"...
  2. J

    Where'd I go wrong?

    Nothing as far as I can tell; you just didn't finish. \frac{-\theta}i = \frac{-\theta i}{i^2} = \frac{-i\theta}{-1} = i\theta ETA: Oops, too slow on the draw.
  3. J

    Directiontal derivative

    I may be wrong, but isn't that just part of the given information? I mean, you're being asked to find the directional derivative, meaning the rate of change of the function at a point in a given direction. So you know the function z=f(x,y), you're given a point P(1,2), and you're given the...
  4. J

    Partial differentiation and changing variables

    You are given that f = f(r) (function of r only). Using the chain rule: \frac{\partial f}{\partial x} = \frac{df}{dr}\cdot\frac{\partial r}{\partial x} (notice that it's df/dr (not partials) because f is a function of r only) Now we need to find \frac{\partial r}{\partial x} ...
  5. J

    Partial differentiation and changing variables

    Use the product rule to find the derivative of this equation: \frac{\partial f}{\partial x} = \frac{x}{r} \cdot \frac{df}{dr}
  6. J

    Why do we learn differentials?

    dy/dx is the instantaneous rate of change of y with respect to x. Since y(x) = x2 is nonlinear, you can't expect dy/dx to equal Δy/Δx
  7. J

    Partial differentiation and changing variables

    Consider the fact that x2 + y2 + z2 = r2 What you've already shown for x also applies to y and z: \frac{\partial f}{\partial x} = \frac{x}{r} \cdot \frac{df}{dr} \frac{\partial f}{\partial y} = \frac{y}{r} \cdot \frac{df}{dr} \frac{\partial f}{\partial z} = \frac{z}{r} \cdot...
  8. J

    Regarding partial fractions

    They probably just defined A and B differently than you did. (They must have said (8x-17)/(x2+x-12) = A/(x-3) + B/(x+4) ). You'll still get the right answer your way.
  9. J

    Calc volume help

    Good grief. I screwed up the algrebra. My second to last step was V = (pi/8)*(a^3 - (a^3/3)) and I managed to get pi*a^3/4 pi*a^3/12 is more like it. Next time I try to help I promise I won't work the answer out on a napkin.
  10. J

    Calc volume help

    Sounds like a funny looking shape; kind of like a loaf of bread or something. Anyway, the base is just a quarter of a circle, so you have x^2+y^2 = a^2 Then you've got semicircles in the z-axis (I'm setting these semicircles' diameters parallel to the x-axis.) To find the volume, I think...