Partial differentiation/extended chain rule question help needed

Hi this is a 1st year UG question, I'm not sure where to start on Q4, any help would be greatly appreciated. Thanks

Well I guess x is the variable as that's changing?

y remains constant

and so does z?

Not sure how I can find the angle, unless I partially differentiate z?

Yes.



Yes.



Well, z is a function of x and of y, and since x is changing, so is z.



Yes, you'll need the partial derivative, and it's hopefully clear from your other answers that it's partial z wrt x that you want.

OK I got 1/100(2x-3y) so far for partial z wrt x, am I supposed to convert this into an angle or something? Also I'm not quite sure about the 'initially ascending or descending' bit of the question

OK, so far.

Worth noting at this stage that you're travelling East, i.e. along the positive x-axis. The y-value is going to be constant and equal to its value at the origin (note it's not zero).

By looking at $\partial z /\partial x$ (which implies y is constant), you're taking a slice through the hill. An elevation plan if you like.

You have effectively reduced the problem to standard one dimensional case.

"ascending/descending", now translates as, is the gradient postive or negative at the origin? I.e. is $\partial z /\partial x$ >0 or < 0?

Regarding the angle. One of the things you should have covered with functions of one variable (which is what you effectively have now), is that the gradient is the tangent of the angle between the curve and the x-axis. And hence you can work out the angle.

OK, so far.

Worth noting at this stage that you're travelling East, i.e. along the positive x-axis. The y-value is going to be constant and equal to its value at the origin (note it's not zero).

By looking at $\partial z /\partial x$ (which implies y is constant), you're taking a slice through the hill. An elevation plan if you like.

You have effectively reduced the problem to standard one dimensional case.

"ascending/descending", now translates as, is the gradient postive or negative at the origin? I.e. is $\partial z /\partial x$ >0 or < 0?

Regarding the angle. One of the things you should have covered with functions of one variable (which is what you effectively have now), is that the gradient is the tangent of the angle between the curve and the x-axis. And hence you can work out the angle.

I got the gradient to be -1 so descending?

I can't quite remember how to use this to work out the angle (mind block)

Agreed.



so, tan(angle of increase)= gradient.

angle of increase = inverse tan (gradient)

And since gradient is negative, you get a negative angle

And, as I'm sure you're aware inverse tan of -1, is -45 degrees.

OK, thanks. For Q5 part a) it asks me to express Fs and Ft in terms of fx, fy, s and t but I'm not quite sure what it's asking me to do, I've got partial derivative x wrt s = t, partial derivative y wrt s = -2t/(s-t)^2, but I don't know how I can work out fx or fy because there's nothing like f(x,y)= x^2 y^3 for example

OK. You can't work out what $f_x,f_y$ are. You need to leave them in, so to speak. You're being asked to express $F_s,F_t$ in terms of them, amongst other things.

Looking at the chain rule $F_s=f_x\frac{\partial x}{\partial s}+f_y\frac{\partial y}{\partial s}$

OK got it, need a bit of help with Q6 a), need to find fx, fy, fz, p, theta, psi, but not sure how to do this as I'm new to spherical coordinates