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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?

Sagot :

abidemiokin

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 [tex]\frac{d\theta}{dt} = \frac{h}{x\sqrt{x^2+y^2} }\frac{dx}{dt}[/tex]

Given the following parameters:

  • Adjacent side = x
  • Opposite side = h
  • The angle of elevation = θ

According to SOH CAH TOA identity;

[tex]tan \theta =\frac{Opposite}{Adjacent}[/tex]

Substitute the given values into the formula

[tex]tan \theta =\frac{h}{x}[/tex]

[tex]\theta= tan^{-1}\frac{h}{x}[/tex]

Differentiate with respect to x

[tex]\frac{d \theta}{dx} = \frac{h}{x\sqrt{x^2+y^2} }[/tex]

According to the chain rule,

[tex]\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}[/tex]

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 [tex]\frac{d\theta}{dt} = \frac{h}{x\sqrt{x^2+y^2} }\frac{dx}{dt}[/tex]

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

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