[tex] \frac{dx}{dt} = 28.2 \: \frac{m}{hr}[/tex]
Step-by-step explanation:
Let y = height of the pole = 35 m (constant)
x = length of the shadow
They are related as
[tex] \tan \theta = \frac{y}{x} [/tex]
or
[tex]x = \frac{y}{ \tan\theta } = y \cot \theta[/tex]
Taking the time derivative of the above expression and keeping in mind that y is constant, we get
[tex] \frac{dx}{dt} = y( - \csc^{2} \theta) \frac{d \theta}{dt} [/tex]
Before we plug in the numbers, let's convert the degree unit into radians:
[tex]40° \times ( \frac{\pi \: rad}{180°}) = \frac{2\pi}{9} \: radians[/tex]
Since the angle is decreasing, then d(theta)/dt is negative. Therefore, the rate at which the shadow is lengthening is
[tex] \frac{dx}{dt} = (35 \: m)( - \csc^{2} \frac{2\pi}{9} )( - \frac{1}{3} \frac{rad}{hr} )[/tex]
or
[tex] \frac{dx}{dt} = 28.2 \: \frac{m}{hr} [/tex]
