How to solve the following question?
If a plane follows the curve $y=f(x)$, and its ground speed is $\frac{dx}{dt} = 500mph$, how fast is the plane going up? How fast is the plane going?
My attempt:
I drew a triangle such that height is $f(x)$ and the base is $x$. The hypotenuse line is z. Since $y = f(x)$, I know that
$$\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}$$.
$$\frac{df}{dt}=\frac{df}{dx}\cdot500$$
Therefore, the speed of plane going up is $\frac{df}{dx}\cdot500$. Next, I use the phytagoras theorem where $z^2 = x^2 + y^2$. This leads to:
$$z^2 = x^2 + f(x)^2$$
Here, I applied implicit differentiation.
$$2z\frac{dz}{dt} = 2x\frac{dx}{dt}+2f(x)\frac{df}{dt}$$
$$\frac{dz}{dt} = \frac{x\frac{dx}{dt}+f(x)\frac{df}{dt}}{z}$$
Substitute the previously obtained "$\frac{df}{dt}$" and predefined $\frac{dx}{dt} = 500$
$$\frac{dz}{dt} = \frac{x\cdot500+f(x)\frac{df}{dx}\cdot500}{z}$$
$$= \frac{500\cdot (x+f(x)\frac{df}{dx})}{z}$$
This is the best I could get. The answer is $500\sqrt{1+(\frac{df}{dx})^2}$ but how to obtain it?