I'm new to derivatives, and I try to solve the following exercise:
A ground camera is being used to film the lift-off of a rocket from its launch pad. The rocket rises vertically so that its vertical distance $y$, in meters, is related to time $t$, in seconds from launch, by $y=10t^2$. Determine the rate of change of the angle of elevation of the camera $10s$ after the launch if the camera is situated horizontally $600m$ from the launch pad.
So my way of thinking is like this: The angle of the camera is related to the rocket's height by the following formula: $$tan(\alpha)=\frac{y(t)}{600meters}=\frac{10t^2}{600}=\frac{1}{60}t^2$$ so $$\alpha=arctan(\frac{1}{60}t^2)$$ So the rate of change of the angle is: $$\frac{d\alpha}{dt}=\frac{d}{dt}arctan(\frac{1}{60}t^2)=\frac{\frac{1}{60}2t}{1+(\frac{1}{60}t^2)^2}=\frac{120t}{t^4 + 3600}$$ So the rate of change of the angle $10s$ after the launch is: $$\alpha'(10seconds)=\frac{120*10}{10^4+3600}=0.088 rad/s$$
But the answer on my solution sheet is $0.9 rad/s$, it's a different order of magnitude. Is there an error in the solution sheet or am I doing something wrong?
Thank you in advance for your help
I think your answer is fine.
Alternative working to check:
$$\tan \alpha = \frac{t^2}{60}$$
$$\sec^2 \alpha \frac{d\alpha}{dt}=\frac{t}{30}$$
$$(\tan^2 \alpha + 1)\frac{d\alpha}{dt}=\frac{t}{30}$$
$$\left(\frac{t^4}{3600}+1 \right)\frac{d\alpha}{dt}=\frac{t}{30}$$
$$(t^4+3600)\frac{d\alpha}{dt}=120t$$
when $t=10$, we do have $$\frac{d\alpha}{dt}=\frac{1200}{13600}=\frac{12}{136}=\frac{3}{34}=0.088$$