Respuesta :
When the balloon, rising at 10 ft/sec, rises to 100 feet, the rate of change of the angle of elevation is 0.05 rad/sec
How can the rate of change of the angle of elevation be found?
The rate at which the balloon rises = 10 ft/sec
Horizontal distance of the balloon from the observer, d = 100 feet
Therefore;
[tex] \frac{dy}{dt} = 10[/tex]
The angle of elevation can be presented as follows;
[tex] tan(\theta) = \frac{y}{100} [/tex]
[tex] y = 100 \times tan(\theta) [/tex]
Therefore;
[tex] \frac{dy}{dt} = \mathbf{ 100\cdot sec^2 \frac{d \theta}{dt}} [/tex]
Which gives;
[tex] \frac{d \theta}{dt}= \frac{ \frac{d y}{dt}}{100} \times cos^2 \theta [/tex]
[tex] \mathbf{ \frac{d \theta}{dt}} = \frac{ 10}{100} \times cos^2 \theta [/tex]
When the balloon is 100 ft. from the ground, we have;
- cos(theta) = 100/(√(100²+100²) = 1/(√2)
Therefore;
[tex] \frac{d \theta}{dt}= \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 [/tex]
[tex] \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 = \frac{ 1}{20} [/tex]
Which gives;
[tex] \frac{d \theta}{dt}=\frac{ 1}{20} = 0.05 [/tex]
- The rate of change of the angle of elevation when the balloon is 100 feet from the ground is 0.05 rad/sec
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