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# Partial differentiation/extended chain rule question help needed watch

1. Hi this is a 1st year UG question, I'm not sure where to start on Q4, any help would be greatly appreciated. Thanks
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2. (Original post by jordanwu)
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 if you're walkiing due East, what variables are changing and what remain the same? And what are you trying to work out?
3. 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?
4. (Original post by jordanwu)
Well I guess x is the variable as that's changing?
Yes.

y remains constant
Yes.

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

Not sure how I can find the angle, unless I partially differentiate 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.
5. (Original post by ghostwalker)
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
6. (Original post by jordanwu)
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 (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 >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.
7. (Original post by ghostwalker)
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 (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 >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)
8. (Original post by ghostwalker)
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 (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 >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.
Also, for Q5 part a) it asks 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
9. (Original post by jordanwu)
I got the gradient to be -1 so descending?
Agreed.

I can't quite remember how to use this to work out the angle (mind block)
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.
10. (Original post by ghostwalker)
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
11. (Original post by jordanwu)
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 are. You need to leave them in, so to speak. You're being asked to express in terms of them, amongst other things.

Looking at the chain rule
12. (Original post by ghostwalker)
OK. You can't work out what are. You need to leave them in, so to speak. You're being asked to express in terms of them, amongst other things.

Looking at the chain rule
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
13. (Original post by jordanwu)
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
This isn't really much different to 5a. You need to decide what x,y,z are as functions of rho,theta,phi. It is actually given in the question, but not explicitly spelt out. Also google spherical coordinates.

I am intentionally not giving the answer, as you need to be more proactive in finding things out, and it's too similar to the previous question.

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