Change of an angle in a triangle

Question

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? enter image description here

This triangle is a model for a spectator looking at an airplane. Does that answer your question? The length of AC changes with 600 km/h — iveqy, Oct 15 '13 at 10:50
Towards A. But that should really matter for more than a -/+ sign, should it? — iveqy, Oct 15 '13 at 11:06
So you mean $B$ and $C$ remain at their place and $A$ is moving at constant speed in direction $AC$? — Michael Hoppe, Oct 15 '13 at 11:10
No C is the airplane moving towards A. The hight above ground (A is a ground level) is BC. B is the point right below the airplane. — iveqy, Oct 15 '13 at 11:12

score 0 · Answer 1 · answered Oct 15 '13 at 11:45

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Either $x(t)=15+600t$ or I didn't get the problem.

answered Oct 15 '13 at 11:45

Michael Hoppe

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You didn't got the problem. I'm interested in at what rate $\theta$ changes. – iveqy Oct 15 '13 at 11:48
If $x(t)$ was $15$, then $\dot x(t)=0$. – Michael Hoppe Oct 15 '13 at 12:28

score 0 · Accepted Answer · answered Oct 15 '13 at 12:42

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.

Change of an angle in a triangle

2 Answers2