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 0.7 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

The answer has to be in rad/s

Answer 2

\[\tan(\theta)=\frac{y}{x}\] is a start

Answer 3

you want \(\theta'\) the rate of change of \(\theta\)

Answer 4

differentiate both sides wrt time get
\[\sec^2(\theta)\theta'=\frac{xy'-yx'}{x^2}\]

Answer 5

quotient rule for the right hand side
now you are told
\[x'=.7\] but damn, you are not told \(y'\) so maybe my method is stupid, or else we have to find it

Answer 6

we can find it, but maybe it is easier to use
\[\cos(\theta)=\frac{x}{10}\] so
\[-\sin(\theta)\theta'=\frac{x'}{10}\] i bet that is simpler

Answer 7

now you are told that \(x'=.7\) so the right hand sides is \(.07\)
we get
\[-\sin(\theta)\theta '=.07\] so now we need \(-\sin(\theta)\)
can you find that?

Answer 8

so we just plug in 0.7 for theta and get -sin(.7) to get our answer?

Answer 9

oh no, .7 is the rate of change of \(x\)

Answer 10

it is not \(\theta\)
in any event, you do not need \(\theta\), you need \(\sin(\theta)\)

Answer 11

when \(x=6\) you know \(y=8\) by pythagoras, making \(\sin(\theta)=\frac{8}{10}=0.8\)

Answer 12

this gives
\[-.8\theta=.07\] so
\[\theta '=-\frac{.07}{.8}=\frac{7}{80}\] whatever that is

Answer 13

oops i meant \(-\frac{7}{80}\)

Answer 14

and now we covert that to radians right?

Answer 15

it is in radians

Answer 16

sine and cosine are functions of numbers, and as such, they correspond to functions of angles only if the angles are measured in radians, not degrees.
so whenever you use sine and cosine, automatically you are working in radians, if you are thinking of angles

Answer 17

oh ok. Yes I get it. Thank you for the thorouugh explanation and steps! And answering my questions

Answer 18

in particular, if you are thinking of sine and cosine as functions of angles measured in degrees, then in fact the derivative of sine is NOT cosine
but no matter, we used them so we are working in radians for sure

Answer 19

yw