Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?

Answer 1

calculus... related rates...
I can probably help, gimme a sec to think :P

Answer 2

L=10?

Answer 3

yes :)
and dL/dt = 0
because the length of the ladder does not change over time

Answer 4

then they gave us
dG/dt = 1.4 when g = 6

and they want us to find d(angle)/dt = ?

Answer 5

I think we can use cos for this

Answer 6

I'm a little confused on how to set up the equation. Related Rates is my enemy.

Answer 7

it was mine too a year ago xD
I just suck at trig right now...
but I think it should go like
cos(angle) = G/L
take the derivative of that, do you know how?

Answer 8

hmm, i think i can try fom here, give me a sec.

Answer 9

mhmm ^_^
chain rule on the left
quociant on the right

then once you get that, solve for d(angle)/dt

Answer 10

nope, I'm stuck.
I used the quotient for the right and got 14/10^2?
and derivative of cos=-sin

Answer 11

don't plug in yet ^_^
so I got -sin(angle) * (d(angle)/dt) = ( L(dG/dt) - G(dL/dt) ) / L^2

Answer 12

oh ok, so can i plug in the numbers now?

Answer 13

i wish this were easier

Answer 14

lol me too
mmm first get d(angle)/dt alone on one side
and know that dL/dt = 0

Answer 15

sooo...
d(angle)/dt = (dG/dt)/L * (-1/sin(angle))

now in order to plug, first find sin(angle) = ?

Answer 16

cow patties, I gtg
sin(angle) = 8/10
dG/dt = 1.4
L = 10
plug it all in
I got d(angle)/dt = -0.175 radians

Answer 17

oh wow thank you!