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Oxford Physics: PAT test discussion watch

1. Oh! So what is the answer then?
3. Ask someone who has done mutlivariable calculus! Idk

Link to the poll as someone requested: https://www.thestudentroom.co.uk/sho....php?t=5033726
4. The answer is actually correct, according to wolframalpha.
5. (Original post by DrSebWilkes)
No, that's technically incorrect. For a department that does maths for a living, they won't have got it wrong intentionally. I know you mean 448/5 t^13 and that's wrong.

If that were right, t=0
But if you say that the area of a circle is pi*r^2 , which could be represented by an intregral involving r, you can differentiate pi*r^2 with respect to r with no problems. Why would that case in question be any different?
6. (Original post by Tomyil12345)
But if you say that the area of a circle is pi*r^2 , which could be represented by an intregral involving r, you can differentiate pi*r^2 with respect to r with no problems. Why would that case in question be any different?
Because t and x are not the same things.

If I ask you to differentiate a constant value, k, with respect to x, what do you get?

well 0! (no, not 1, I mean zero).

So if I then integrate x to a constant value, t, then my area under the curve is equal to whatever the integral of x is, with the values of "t" subbed in. t could be "5" ... or -555555 or 55555555555555555555555555555

But what you can't do is integrate to a variable - that's not strictly allowed using a different variable without some other relationship involved.

Given no other relationship between the variables t and x, my teacher says this cannot be solved unless it equals 0 (using A-level methods, at which he mentioned this multivariable calculus rule)

The answer is actually correct, according to wolframalpha.
*sigh* wolfram alpha isn't treating the numbers properly; it's a machine being programmed to follow blindly whatever you input; of course it will spit that out.
7. Nope. Wolframalpha differentiated the integral partially wrt variable t. I think the answer is correct.

You say that the method I used is wrong; maybe. But u also say that the answer obtained is wrong.... Well, could you please get the right answer then? If you don't know what the correct answer will be using the "correct method", how can you be sure that this answer isn't? (even though the wrong method is used)
8. (Original post by Tomyil12345)
We equally have a bunch of questions on maths exams were you have to solve for the variable in a integration, if t is just a (real)number you’re not doing anything illegal, you do however have to assume that t is independent of x i think
Then if it is independent of x, it is still either a variable or constant; if t is a constant, d((f(x))/dt = 0

if it is a variable, then idk how to solve it - do first year maths to find out how.

When do you it in exams, you are integrating to a fixed (constant) point.

EG the work done by gravity to move an object from infinity to a distance "r" away from the other object. That "r" isn't changing? It's a set value - sure we can change that r, so r is variable in the sense it isn't a measurement. So yeah we can change "r" by giving it a value, but r only changes in a certain way with respect to r; without any other relationship, r doesn't effect heat or something does it? So you can't do dr/dH ... that's not how it works

Nope. Wolframalpha differentiated the integral partially wrt variable t. I think the answer is correct.

You say that the method I used is wrong; maybe. But u also say that the answer obtained is wrong.... Well, could you please get the right answer then? If you don't know what the correct answer will be using the "correct method", how can you be sure that this answer isn't? (even though the wrong method is used)

I can't no .. but if all else fails, here is a 2nd year university student telling me how what we did is wrong. Maybe we will get the marks but it would be highly irregular for quasi-mathematicians in Oxford to get this wrong.

Attachment 701668
9. (Original post by DrSebWilkes)
Because t and x are not the same things.

If I ask you to differentiate a constant value, k, with respect to x, what do you get?

well 0! (no, not 1, I mean zero).

So if I then integrate x to a constant value, t, then my area under the curve is equal to whatever the integral of x is, with the values of "t" subbed in. t could be "5" ... or -555555 or 55555555555555555555555555555

But what you can't do is integrate to a variable - that's not strictly allowed using a different variable without some other relationship involved.

Given no other relationship between the variables t and x, my teacher says this cannot be solved unless it equals 0 (using A-level methods, at which he mentioned this multivariable calculus rule)

*sigh* wolfram alpha isn't treating the numbers properly; it's a machine being programmed to follow blindly whatever you input; of course it will spit that out.
t was a bound of the integral, thus if t represents any value it should be a real number( i think) you can integrat functions with variables representing real numbers with no problems
10. Btw... I read a little bit about the Leibniz rule. And here is a special case of it where it allows us to do what we did to solve the problem (the 3rd big equation line):

So if the limits are not functions of x.... Then what we did is correct. Hence, the answer is correct.
11. (Original post by DrSebWilkes)
Then if it is independent of x, it is still either a variable or constant; if t is a constant, d((f(x))/dt = 0

if it is a variable, then idk how to solve it - do first year maths to find out how.

When do you it in exams, you are integrating to a fixed (constant) point.

EG the work done by gravity to move an object from infinity to a distance "r" away from the other object. That "r" isn't changing? It's a set value - sure we can change that r, so r is variable in the sense it isn't a measurement. So yeah we can change "r" by giving it a value, but r only changes in a certain way with respect to r; without any other relationship, r doesn't effect heat or something does it? So you can't do dr/dH ... that's not how it works
If you differentiate a constant you get zero, but the integral solution was t^13 someting.. if you can’t differentiate this because t is a variable you couldn’t differentiate y=x^2 either
Btw... I read a little bit about the Leibniz rule. And here is a special case of it where it allows us to do what we did to solve the problem (the 3rd big equation line):

So if the limits are not functions of x.... Then what we did is correct. Hence, the answer is correct.
Woah I really don't see how we are right? Can't you see it uses partial derivatives?!
13. does anyone remeber the parametric equations?
14. (Original post by Tomyil12345)
If you differentiate a constant you get zero, but the integral solution was t^13 someting.. if you can’t differentiate this becouse t is a variable you couldn’t differentiate y=x^2 either
Obviously you can; that is the fundamental idea of calculus.

If y is related to x, then we change see how y changes when we change x by a small amount dx. Hence, of course we can differentiate y=x^2, and y=38278.8282828822882 * x^-2288282.2

integrating to a limit, at our level, means it is a CONSTANT. If we then differentiate the expression, we have just differentiated a constant!
15. Tomyil... What he is saying is, if the limit is a CONSTANT.... Then the 't' is constant. So if u Differentiate it, you'll get 0 regardless of what u Differentiate it wrt.

However, since this is multivariable Calc,. While integrating the function wrt x, I think the t variable is TREATED as constant, and while it is being differentiated wrt t, t is treated as variable and X, if present, is constant.
16. (Original post by DrSebWilkes)
Woah I really don't see how we are right? Can't you see it uses partial derivatives?!
Yes. Partial differentiation just treats the other variables as constant while being differentiated. So as per the special case, interchange the integral and derivative. Differentiate partially wrt t by treating x as constant, and then integrate it. You'll get the answer we got first.
17. Anyone remember the parametric equations
18. (Original post by DrSebWilkes)
Obviously you can; that is the fundamental idea of calculus.

If y is related to x, then we change see how y changes when we change x by a small amount dx. Hence, of course we can differentiate y=x^2, and y=38278.8282828822882 * x^-2288282.2

integrating to a limit, at our level, means it is a CONSTANT. If we then differentiate the expression, we have just differentiated a constant!
Yes exacly, now if you ignore that you got t^13 by an integral you can diff. this with respect to t. Since t is a variable independent of x( otherwise they would have to state this) it doesn’t matter that you obtained that which you are diffirentiating by an integral
Tomyil... What he is saying is, if the limit is a CONSTANT.... Then the 't' is constant. So if u Differentiate it, you'll get 0 regardless of what u Differentiate it wrt.

However, since this is multivariable Calc,. While integrating the function wrt x, I think the t variable is TREATED as constant, and while it is being differentiated wrt t, t is treated as variable and X, if present, is constant.
Exacly
20. (Original post by igcsesareeasy)
Anyone remember the parametric equations
X = a(wt - sin(wt))
Y = a(sqrt(3) - 2cos(wt))

(95% sure)

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