14) A spot Light is on the ground \( (20 \mathrm{ft} \) away from a wall and 6 FT ) person is walking towards the wall at a rate of \( 2.5 \mathrm{tT} / \mathrm{sec} \). How fast is the height of the shadow changing when the person is 8 fT from the wall? Is the shadow increasing or decreasing in height at this Time?

To solve this problem, we can use related rates in calculus. Let's define the variables involved: - Let \( h \) be the height of the shadow. - Let \( x \) be the distance of the person from the wall. - The distance from the spotlight to the wall is a constant \( 20 \) ft. - The height of the person is \( 6 \) ft. - The person is walking towards the wall at a rate of \( \frac{dx}{dt} = -2.5 \) ft/sec (negative because \( x \) is decreasing). ### Step 1: Set up the relationship Using similar triangles, we can set up the following relationship: \[ \frac{h}{20} = \frac{6}{x} \] From this, we can express \( h \) in terms of \( x \): \[ h = \frac{6}{20} x = \frac{3}{10} x \] ### Step 2: Differentiate with respect to time Now, we differentiate both sides with respect to time \( t \): \[ \frac{dh}{dt} = \frac{3}{10} \frac{dx}{dt} \] ### Step 3: Substitute known values We need to find \( \frac{dh}{dt} \) when the person is \( 8 \) ft from the wall. At that moment, \( x = 8 \) ft. We also know \( \frac{dx}{dt} = -2.5 \) ft/sec. Now we can substitute \( \frac{dx}{dt} \) into the differentiated equation: \[ \frac{dh}{dt} = \frac{3}{10} (-2.5) \] ### Step 4: Calculate \( \frac{dh}{dt} \) Now, let's calculate \( \frac{dh}{dt} \): \[ \frac{dh}{dt} = \frac{3}{10} \times -2.5 = -\frac{7.5}{10} = -0.75 \text{ ft/sec} \] ### Conclusion The height of the shadow is changing at a rate of \( -0.75 \) ft/sec when the person is \( 8 \) ft from the wall. Since the rate is negative, the shadow is decreasing in height at this time.