Mathematics: if the angle of elevation of the sun is 40 degrees, and is decreasing 1/3 radians/hour how fast is the shadow of a 35m tall pole lengthening?

B) 90 degrees clockwise rotation...

if the angle of elevation of the sun is 40 degrees,

karkarntx

13.07.2021

$\frac{dx}{dt} = 28.2 \: \frac{m}{hr}$

Step-by-step explanation:

Let y = height of the pole = 35 m (constant)

x = length of the shadow

They are related as

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

$x = \frac{y}{ \tan\theta } = y \cot \theta$

Taking the time derivative of the above expression and keeping in mind that y is constant, we get

$\frac{dx}{dt} = y( - \csc^{2} \theta) \frac{d \theta}{dt}$

Before we plug in the numbers, let's convert the degree unit into radians:

$40° \times ( \frac{\pi \: rad}{180°}) = \frac{2\pi}{9} \: radians$

Since the angle is decreasing, then d(theta)/dt is negative. Therefore, the rate at which the shadow is lengthening is

$\frac{dx}{dt} = (35 \: m)( - \csc^{2} \frac{2\pi}{9} )( - \frac{1}{3} \frac{rad}{hr} )$

$\frac{dx}{dt} = 28.2 \: \frac{m}{hr}$

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