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

8 0

2 years ago

Solve this equation:

lilavasa [31]

Answer:

The value of x = 6

Step-by-step explanation:

8x - 13 = 11 + 5x

collect like terms on either sides

8x - 5x = 11 + 13

3x = 24

x = 24/3

x = 6

5 0

3 years ago