Question

Related Rates. Help needed.

Answer 1

0.031 radians/second

Explanation:

A balloon is at a point on the ground 50 m from an observer. If θ is the angle of elevation from the observer to the balloon, and y is the height of the balloon above the ground, the equation that relates θ to y can be found by using the tangent ratio:

`\tan \theta = (y)/(50)`

`y=50\tan \theta`

To find the rate of change of the angle of elevation (θ) from the observer to the balloon when the balloon is y meters above the ground, we need to find dθ/dt. To do this, we can use the chain rule:

`\frac{\text{d}\theta}{\text{d}t}=\frac{\text{d}\theta}{\text{d}y} * \frac{\text{d}y}{\text{d}t}`

To find dθ/dy, differentiate y = 50 tan θ:

`\frac{\text{d}y}{\text{d}\theta}=50\sec^2\theta`

Use the identity sec²θ = 1 + tan²θ:

`\frac{\text{d}y}{\text{d}\theta}=50(1+\tan^2\theta)`

`\frac{\text{d}y}{\text{d}\theta}=50+50\tan^2\theta`

Therefore:

`\frac{\text{d}\theta}{\text{d}y}=(1)/(50+50\tan^2\theta)`

Given the balloon is rising at a rate of 2 m/s:

`\frac{\text{d}y}{\text{d}t}=2\; \sf m/s`

Therefore:

`\frac{\text{d}\theta}{\text{d}t}=\frac{\text{d}\theta}{\text{d}y} * \frac{\text{d}y}{\text{d}t}`

`\frac{\text{d}\theta}{\text{d}t}=(1)/(50+50\tan^2\theta)* 2`

`\frac{\text{d}\theta}{\text{d}t}=(2)/(50+50\tan^2\theta)`

`\frac{\text{d}\theta}{\text{d}t}=(1)/(25+25\tan^2\theta)`

To find the rate of change of the angle of elevation (θ) from the observer to the balloon when the balloon is 27 m above the ground, first find the value of tan θ when y = 27 by substituting y = 27 into y = 50 tan θ:

`27=50\tan \theta`

`\tan \theta=(27)/(50)`

Now, substitute the value of tan θ into dθ/dt:

`\frac{\text{d}\theta}{\text{d}t}=(1)/(25+25\left((27)/(50)\right)^2)`

`\frac{\text{d}\theta}{\text{d}t}=(1)/(25+25\left((729)/(2500)\right))`

`\frac{\text{d}\theta}{\text{d}t}=(1)/(25+(729)/(100))`

`\frac{\text{d}\theta}{\text{d}t}=(1)/((3229)/(100))`

`\frac{\text{d}\theta}{\text{d}t}=(100)/(3229)`

`\frac{\text{d}\theta}{\text{d}t}=0.03096934035...`

`\frac{\text{d}\theta}{\text{d}t}=0.031\; \sf radians/second\;(3\;d.p.)`

Therefore, the rate of change of the angle of elevation from the observer to the balloon when the balloon is 27 meters above the ground is 0.031 radians/second (rounded to three decimal places).