I've a triangle ABC. Where AC is the hypotenuse and the angle ABC is 90 degress.
AB is $15 km$ and changes with a speed of $600 km/h$. BC is $5 km$ and changes with a speed of $0 km/h$. At what speed changes the angle CAB in terms of $rad/h?$
I call AB for $x(t)$ and BC for $y(t)$. Then I know:
$$x(t) = 15 km$$ $$x'(t) = 600 km/h$$ $$y(t) = 5 km$$ $$y'(t) = 0 km/h$$
To find the angle CAB (now called $\theta$):
$$ tan(\theta) = \frac{y(t)}{x(t)} \leftrightarrow \theta = arctan\left(\frac{y(t)}{x(t)}\right)$$.
I derivate this to get the change of $\theta$ in $rad/h$.
$$ \frac{1}{1+\left(\large\frac{y(t)}{x(t)}\right)^2} \times \left(\frac{y'(t)x(t)-y(t)x'(t)}{x(t)^2}\right) = 0.0075~rad/h$$
I know that the answer should be $12~rad/h$ so somewhere I'm wrong. But where?
2 Answers
$\begingroup$For clarification here, $x(0)=15, y(0)=5$ is what is meant here. OP, you need to write it like this, otherwise things get real messy.
As for an answer:
We know that: $$ x(0)=15\\ y(0)=5\\ \frac{dx}{dt}=x'(t)=600\\\frac{dy}{dt}=y'(t)=0 $$
So from this we can get $x(t)=600t+15$ and $y(t)=5$. We're aiming for $\frac{d\theta}{dt}$, and to get this we can multiply $\frac{dx}{dt}$ and $\frac{d\theta}{dx}$, as the $dx$ cancels out. We already have $\frac{dx}{dt}$, so to get $\frac{d\theta}{dx}$: $$\begin{align*} tan(\theta) & = \frac{y(t)}{x(t)} \\ & = \frac{5}{x} \\ \end{align*} $$
We'll treat x as a variable by itself for the moment:
$$\theta = arctan(\frac{5}{x})\\ \begin{align*}\frac{d\theta}{dx} & = \frac{-5}{x^2}\cdot\frac{1}{1+(\frac{5}{x})^2}\\ & = \frac{-5}{x^2}\cdot\frac{1}{\frac{1}{x^2}(x^2+25)}\\ & = \frac{-5x^2}{x^2(x^2+25)}\\ & = \frac{-5}{x^2+25} \end{align*}$$
so substituting into $\frac{dx}{dt}\cdot\frac{d\theta}{dx}=\frac{d\theta}{dt}$, we get:
$$\begin{align*} \frac{d\theta}{dt} & = \frac{-5\cdot600}{x^2+25}\\ & = \frac{-3000}{(600t+15)^2+25}\\ & = \frac{-3000}{360000t^2+18000t+225+25}\\ & = \frac{-3000}{360000t^2+18000t+250}\\ & = \frac{-60}{7200t^2+3600t+5} \end{align*}$$
As $t=0$, we can substitute back into our equation for $\frac{d\theta}{dt}$ to get:
$$\begin{align*} \frac{d\theta}{dt} & = \frac{-60}{5}\\ & = -12\\ \end{align*}$$
So at $t=0$, $\theta$ is decreasing at 12 rad/h.
$\endgroup$ $\begingroup$Either $x(t)=15+600t$ or I didn't get the problem.
$\endgroup$ 2
