A spotlight on the ground is shining on a wall 24\text{m} away. If a woman 2\text{m} tall walks from the spotlight toward the building at a speed of 0.6\text{m/s}, how fast is the length of her shadow on the building decreasing when she is 2\text{m} from the building?
To find how fast the length of her shadow on the building is decreasing, we need to find the rate of change of the length of the shadow with respect to time. Let's denote the length of the shadow as "s" (in meters) and time as "t" (in seconds).
We are given that the woman is 2 meters tall, and the spotlight is 24 meters away from the wall. Therefore, when the woman is 2 meters away from the building, the distance from the spotlight to the building is 24 - 2 = 22 meters.
We can set up a similar triangles relationship between the woman, her shadow, and the distance from the spotlight to the building:
\(\frac{s}{\text{woman's height}} = \frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\)
Differentiating both sides with respect to time, we get:
\(\frac{ds}{dt} \cdot \frac{1}{\text{woman's height}} = \frac{d}{dt}\left(\frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\right)\)
We are asked to find \(\frac{ds}{dt}\) when the woman is 2 meters from the building, so we need to find \(\frac{d}{dt}\left(\frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\right)\) at that point.
Differentiating the expression inside the parentheses using the quotient rule, we get:
\(\frac{d}{dt}\left(\frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\right) = \frac{\frac{d}{dt}(\text{distance from spotlight to building}) \cdot \text{distance from spotlight to woman} - \text{distance from spotlight to building} \cdot \frac{d}{dt}(\text{distance from spotlight to woman})}{(\text{distance from spotlight to woman})^2}\)
Now let's find the derivatives of the distances:
The distance from the spotlight to the woman is decreasing at a rate of 0.6 m/s since the woman is moving towards the building at a speed of 0.6 m/s:
\(\frac{d}{dt}(\text{distance from spotlight to woman}) = -0.6\) (m/s)
The distance from the spotlight to the building is constant at 24 meters, so its derivative is 0:
\(\frac{d}{dt}(\text{distance from spotlight to building}) = 0\) (m/s)
Substituting these values into the expression, we get:
\(\frac{d}{dt}\left(\frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\right) = \frac{0 \cdot \text{distance from spotlight to woman} - 24 \cdot (-0.6)}{(\text{distance from s potlight to woman})^2}\)
Simplifying further:
\(\frac{d}{dt}\left(\frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\right) = \frac{14.4}{\text{distance from spotlight to woman}^2}\)
We want to find \(\frac{ds}{dt}\) when the woman is 2 meters away from the building, so substitute this value into the expression:
\(\frac{d}{dt}\left(\frac{\text{distance from spotlight to building}}{\text{distance from spotlight to woman}}\right) = \frac{14.4}{2^2} = \frac{14.4}{4} = 3.6\) (m/s)
Therefore, when the woman is 2 meters from the building, the length of her shadow on the building is decreasing at a rate of 3.6 m/s.
when she is x meters from the light, and her shadow has height y, then
2/x = y/24
-2/x^2 dx/dt = 1/24 dy/dt
When she is 2m from the building (22 m from the light),
-2/22^2 * 0.6 = 1/24 dy/dt
dy/dt = -0.06 m/s