I'm having trouble finding a formula to relate the time with the angle. This is the question:
A $5$ meter ladder rests against a vertical wall at an angle $\theta$ with the horizontal surface as shown below.
The bottom of the ladder is $1$ meter from the wall. If the bottom of the ladder slides away from the wall at a constant rate of $0.5m/s$, at what rate will $\theta$ decrease after $4$ seconds?
My attempt:
$$ \frac{dx}{dt}=0.5m/s $$ $$ \frac{d\theta}{dt} ? $$ $$ x=1m $$ $$ t=4s $$ $$ cos\theta=\frac{A}{H}=\frac{x}{5} $$ $$ \theta=cos^{-1}(\frac{x}{5}) $$ $$ d\theta = -\frac{1}{\sqrt{1-\frac{x^2}{25}}}dx $$ $$ \frac{d\theta}{dx} = -\frac{1}{\sqrt{1-\frac{x^2}{25}}} $$ $$ \frac{d\theta}{dt}=\frac{d\theta}{dx}\frac{dx}{dt} $$ $$ \frac{d\theta}{dt}=(-\frac{1}{\sqrt{1-\frac{x^2}{25}}})(0.5) $$ $$ \frac{d\theta}{dt}=(-\frac{1}{\sqrt{1-\frac{1}{25}}})(0.5) $$ $$ \frac{d\theta}{dt}=-0.5103m/s $$
Decreasing at a rate of $0.5103m/s$.
What about the $t=4s$? I'm not sure how to use that because there's no formula that I've learnt that has time and angles in it.
Thanks
You are almost there, except that you have substituted the wrong value for $x$: You want the derivative at $t=4$, but you have it in terms of $x$; when $t=4$, the value of $x$ is $1 + 0.5 \times 4 = 3$. So plug $x=3$ into your expression for the derivative instead.
Also, your expression for $\frac{dx}{dt}$ has a problem: while you are sliding away from the wall, your value of $x$ is in fact increasing, so you shouldn't put a negative sign in front of the rate.(I see that's been corrected.)Think of all your variables as a system: when $t$ is some value, then $x$ and $\theta$ take some corresponding value. You've started with $x=1$ when $t=0$; when $t$ changes, so does $x$.