Question

Could someone explain this Calc problem to me pwease? :3

Answer 1

first use trigonometry to write a formula for x in terms of theta

Answer 2

so
sin(theta) = x/5 ?

Answer 3

differentiate with respect to time

Answer 4

right, and I'd recommend getting that 5 on the other side, though you can do that later if you want

either way, differentiate both sides with respect to time

Answer 5

how would I "differentiate with respect to time"? like do dy/dt ?
but I don't have y's or t's ?

Answer 6

you have f(theta) on one side, and f(x) on the other side
use the chain rule

Answer 7

yeah d/dt both sides

Answer 8

d/dt f(u) = df/du * du/dt

Answer 9

huh?
so d/dt sin(theta) = d/dt (1/5) * dx/dt ?

Answer 10

erm... sort of
let theta=u
on the left then we have
d/dt sin(u)=d/du sin(u) * du/dt

what is d/du sin(u) ?

Answer 11

mmm 3?

Answer 12

nope, what is d/dx sin(x) ?

Answer 13

cos(x)

Answer 14

right, so d/du sin(u) is...?

Answer 15

cos(u) ?

Answer 16

good, then using the chain rule we have
d/dt[sin(u)]=cos(u)*du/dt
now remember that u is theta, so what is du/dt?

Answer 17

"theta increases at a constant rate of 3 radians per minute"
so 3 ?

Answer 18

excellent :)
now for the right side
what is d/dt (x/5) ?

Answer 19

(1/5)(dx/dt) ?

Answer 20

great, and we want dx/dt, so we're almost there!
now get that 5 on the other side and what do we have?

Answer 21

mm
dx/dt = 15cos(u)

Answer 22

great, so clearly the one thing we need now is cos(u)
any ideas how we know to find that?

Answer 23

oh right
x was given as 3
so cos(u) = 3/5

Answer 24

careful, look where theta is!
sin(u)=3/5, so cos(u)=?

Answer 25

no wait...

Answer 26

erm...
root ( 5^2 - 3^2) = 4
so cos(u) = 4/5 ?

Answer 27

bingo, so dx/dt is....?

Answer 28

12 :)
thank you!

Answer 29

very welcome!