Question

any one who can help me with calc 3

Answer 1

is calc 3 differential equations?

Answer 2

F=cos(pi y i)-pi x sin(pi y)j ,,is this a gradient?

Answer 3

Hmm it's a bit hard to read, for me.

\[ F= \cos(\pi y)i -(\pi x \sin (\pi y)j \]
? I hardly doubt that I got it correct, so maybe you can draw it or upload a picture/copy of the problem?

Answer 4

you got it right

Answer 5

Hehe, the power of \( \LaTeX\), okay let me take a look at it.

Answer 6

To verify if this is an exact equation you want to check if \[\Large M_y=N_x\] You might have seen that before in exact equations.

Answer 7

Excuse me if I did empathize a bit too much with the exact equation terminology, it's how I recall it. Basically you use the same methods of checking, if the statements hold, you have a gradient field.

Answer 8

you right,,,but i want to verify for that particular question,,

Answer 9

is it gradient or not,,i got it's not but the answer turns to be it's gradient in my book

Answer 10

Good, so maybe the following will do the trick.

\[ \Large F= \underbrace{\cos(\pi y)}_Mi -\underbrace{(\pi x \sin (\pi y)}_Nj \]

Answer 11

yes it is a gradient field, differentiate the M term with respect to y, partial derivative, then take the partial with respect to x of the second term.

Answer 12

cos pi y gives me - pi( sin pi Y)

Answer 13

am i right

Answer 14

exactly \[\Large M_y=- \pi \sin (\pi y) \]

Answer 15

the second piece didn't gave me the same answer though,,,any clue?

Answer 16

\[\Large N= -\pi x \sin (\pi y) \]

( !! I did distribute the minus sign infront of the \(j\) component already))

Answer 17

Now you want to take the partial of that with respect to x, so all the terms that include a y are just a constant.

Answer 18

in fact \( \sin (\pi y) \) is a constant.

Answer 19

thanks

Answer 20

really apperitiate

Answer 21

very welcome, hope that helped.

Answer 22

yep it did