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=hx. Which of the following best describes the relationship between dθdt, the rate of change of θ with respect to time, and dxdt, the rate of change of x with respect to time?

Answer 1

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 is $\frac{d\theta}{dt} = \frac{h}{x\sqrt{x^2+y^2} }\frac{dx}{dt}$

Given the following parameters:

• Adjacent side = x

• Opposite side = h

• The angle of elevation = θ

According to SOH CAH TOA identity;

$tan \theta =\frac{Opposite}{Adjacent}$

Substitute the given values into the formula

$tan \theta =\frac{h}{x}$

$\theta= tan^{-1}\frac{h}{x}$

Differentiate with respect to x

$\frac{d \theta}{dx} = \frac{h}{x\sqrt{x^2+y^2} }$

According to the chain rule,

$\frac{d\theta}{dt} =\frac{d\theta}{dx} \times \frac{dx}{dt}\\\frac{d\theta}{dt} = \frac{h}{x\sqrt{x^2+y^2} }\frac{dx}{dt}$

Hence 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 is $\frac{d\theta}{dt} = \frac{h}{x\sqrt{x^2+y^2} }\frac{dx}{dt}$

Learn more on the rate of change here: brainly.com/question/11883878