Let O denote the origin of the axis of co-ordinates, and let C denote the part of the parabola $y=x^2$ which lies in the first quadrant. A particle P starts at O and moves along C in such a way that its distance from the y-axis is increasing at a rate of 3 units per second. Let Q denote the point where the tangent to the parabola at the point P intersects the x-axis, and let R denote the foot of the perpendicular from P to the x-axis. Let $\phi$ denote the angle (in radians) subtended at the vertex Q of the triangle PQR. How fast is $\phi$ changing, when P is 4 units away from the y-axis?
My attempt at a solution:
We are given $\frac{dx}{dt}$ $=3$ units/second. We'll use $P(u,u^2)$. We want $\frac{d\phi}{dt}$ when $u=4$.
When P is 4 units away from the y-axis, we have $P(4,16)$ and $R(4,0)$. I found the x-intercept of Q by writing the following equations:
$$y-u^2=2u(x-u)$$
$$0-16=2(4)(x-4)\Rightarrow x=2$$
So Q lies on $(2,0)$. We now have a triangle with side lengths $||PQ||=\sqrt{20}, ||PR||=4, ||QR||=2$.
To find $\phi$: $$tan(\phi)=\frac{||PR||}{||QR||}=\frac{16}{2}=8\Rightarrow\frac{d\phi}{dx}=arctan(8)$$ $$\frac{d\phi}{dt}=\frac{d\phi}{dx}\cdot\frac{dx}{dt}=arctan(8)\cdot3$$
Am I even approaching this correctly? I'm really not sure if that will work or if I should be using $||PQ||$. We were also supposed to find the area of the triangle at this instant and I found it to be 48, but for some reason I'm unsure about the angle part and if this method will still give me the correct value over a change in time. Thank you!
You got something wrong, $\phi =\arctan (8)$ but not $\frac{d\phi}{dx}
Note that tangent of the angle is simply $\tan \phi=\frac{dy}{dx}=2x$ because the slope of the line $PQ$ is the same as the derivative of the curve at that point. Differentiate both sides w.r.t. $x$
$$\sec^2 \phi \frac{d\phi}{dx}=2$$
$$\frac{d\phi}{dt}=2\cos^2 \phi \frac{dx}{dt}$$