Question

a right triangle has base x meters and height h meters, where h is constant and x changes with respect to time t, measured in seconds. the angle θ, measured in radians, is defined by tanθ = h/x. which of the following best describes the relationship between dθ/dt, the rate of change of θ with respect to time and dx/dt the rate of change of x with respect to time?
a dθ/dt = (h/(x² + h²)) dx/dt radians per second
b dθ/dt = (-h/(x² + h²)) dx/dt radians per second
c dθ/dt = (h/(x√(x² + h²))) dx/dt radians per second
d dθ/dt = (-h/(x√(x² + h²))) dx/dt radians per second

Explanation:

Step1: Differentiate the equation \(\tan\theta=\frac{h}{x}\) with respect to \(t\)

We know that \(\frac{d}{dt}(\tan\theta)=\frac{d}{dt}(\frac{h}{x})\).
Using the chain - rule, \(\frac{d}{dt}(\tan\theta)=\sec^{2}\theta\frac{d\theta}{dt}\).
Using the quotient rule, \(\frac{d}{dt}(\frac{h}{x})=\frac{0\times x - h\times\frac{dx}{dt}}{x^{2}}=-\frac{h}{x^{2}}\frac{dx}{dt}\).
So, \(\sec^{2}\theta\frac{d\theta}{dt}=-\frac{h}{x^{2}}\frac{dx}{dt}\).

Step2: Express \(\sec^{2}\theta\) in terms of \(x\) and \(h\)

Since \(\tan\theta=\frac{h}{x}\), and \(\sec^{2}\theta = 1+\tan^{2}\theta\), then \(\sec^{2}\theta=1 + (\frac{h}{x})^{2}=\frac{x^{2}+h^{2}}{x^{2}}\).
Substitute \(\sec^{2}\theta=\frac{x^{2}+h^{2}}{x^{2}}\) into \(\sec^{2}\theta\frac{d\theta}{dt}=-\frac{h}{x^{2}}\frac{dx}{dt}\).
We get \(\frac{x^{2}+h^{2}}{x^{2}}\frac{d\theta}{dt}=-\frac{h}{x^{2}}\frac{dx}{dt}\).
Then \(\frac{d\theta}{dt}=-\frac{h}{x^{2}+h^{2}}\frac{dx}{dt}\).

Answer:

B. \(\frac{d\theta}{dt}=(\frac{-h}{x^{2}+h^{2}})\frac{dx}{dt}\) radians per second