A hot air balloon lifts off from the ground rising vertically. From 100 feet away, a 5’ woman tracks the path of the balloon. When her sightline with the balloon makes a \(45^\circ\) angle with the horizontal, she notes the angle is increasing at about \(5^\circ\) per minute.
(a) What is the elevation of the balloon?
(b) How fast is it rising?
Let \(h\) be the elevation of the balloon above the ground (in feet).
Let \(d\) be the horizontal distance from the woman to the balloon (in feet).
Given that the woman is 5 feet tall and is 100 feet away from the balloon, we can use trigonometry to find \(h\).
The tangent function relates the angle \(\theta\) to the ratio of the opposite side (height \(h\)) to the adjacent side (distance \(d\)):
\[ \tan(\theta) = \frac{h}{d} \]
At the moment when the angle \(\theta\) is \(45^\circ\), we have:
\[ \tan(45^\circ) = \frac{h}{100} \]
Solving for \(h\):
\[ h = 100 \times \tan(45^\circ) \]
\[ h = 100 \times 1 \]
\[ h = 100 \text{ feet} \]
So, the elevation of the balloon is \(100\) feet.
Let \(\frac{dh}{dt}\) be the rate at which the balloon is rising in feet per minute.
Given that the angle \(\theta\) is increasing at about \(5^\circ\) per minute:
\[ \frac{d\theta}{dt} = 5^\circ \text{ per minute} \]
We know that when \(\theta = 45^\circ\), \(h = 100\) feet.
Use tangent:
\[ \tan(\theta) = \frac{h}{100} \]
Differentiate both sides with respect to time \(t\):
\[ \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{100} \frac{dh}{dt} \]
At \(\theta = 45^\circ\), \(\sec^2(45^\circ) = 1\), so:
\[ \frac{1}{100} \times 5^\circ = \frac{1}{100} \frac{dh}{dt} \]
\[ \frac{1}{100} \times 5^\circ = \frac{1}{100} \frac{dh}{dt} \]
\[ \frac{1}{100} \times 5 = \frac{dh}{dt} \]
\[ \frac{dh}{dt} = 0.05 \text{ feet per minute} \]
The balloon is rising at a rate of \(0.05\) feet per minute.