A hot air ballon lifts off the earth 300m away from an observer and rises straight up at a rate of 80m/min. At what rate is the angle of inclination of the observers line of sight increasing at the time the balloon is 300, above the ground?
when the balloon is at height h,
tanθ = h/300
so,
sec^2θ dθ/dt = 1/300 dh/dt
you know that when h=300, θ=π/4.
You know dh/dt = 80, so just solve for dθ/dt
To find the rate at which the angle of inclination of the observer's line of sight is increasing, we can use the concepts of similar triangles and trigonometry.
Let's define a few variables:
- Let \(x\) represent the height of the balloon above the ground at any given time.
- Let \(\theta\) represent the angle of inclination of the observer's line of sight.
- Let \(L\) represent the horizontal distance between the observer and the balloon.
We know that the balloon is 300m away from the observer initially, so \(L = 300\).
Using similar triangles, we can set up the following relationship:
\[\frac{x}{L} = \tan(\theta)\]
To find the rate of change of the angle of inclination with respect to time, we need to take the derivative of both sides of the equation.
\[\frac{d}{dt}\left(\frac{x}{L}\right) = \frac{d}{dt}\left(\tan(\theta)\right)\]
Now, let's differentiate each side step by step:
\[\frac{1}{L}\cdot \frac{dx}{dt} = \sec^2(\theta) \cdot \frac{d\theta}{dt}\]
Since \(\frac{dx}{dt}\) is the rate at which the balloon is rising, it is given as 80m/min.
\[\frac{1}{300} \cdot 80 = \sec^2(\theta) \cdot \frac{d\theta}{dt}\]
To find the value of \(\theta\), we can use the Pythagorean theorem:
\[L^2 + x^2 = (\text{distance of observer to balloon})^2\]
\[300^2 + x^2 = 300^2\]
\[x^2 = 0\]
Therefore, \(x = 0\). This means that the height of the balloon above the ground is 0 when it is directly above the observer.
Now, we can substitute \(x = 0\) into our equation:
\[\frac{1}{300} \cdot 80 = \sec^2(\theta) \cdot \frac{d\theta}{dt}\]
Simplifying:
\[\frac{4}{15} = \sec^2(\theta) \cdot \frac{d\theta}{dt}\]
Finally, we can solve for \(\frac{d\theta}{dt}\):
\[\frac{d\theta}{dt} = \frac{4}{15 \cdot \sec^2(\theta)}\]
Since we want to find the value of \(\frac{d\theta}{dt}\) when the balloon is 300m above the ground (\(x = 300\)), we need to find the corresponding value of \(\theta\).
Using the trigonometric identity \(\tan(\theta) = \frac{x}{L}\), we can rearrange and solve for \(\theta\):
\[\tan(\theta) = \frac{x}{L}\]
\[\tan(\theta) = \frac{300}{300}\]
\[\tan(\theta) = 1\]
Taking the inverse tangent of both sides:
\[\theta = \tan^{-1}(1)\]
\[\theta = 45^\circ\]
Now, we can substitute this value of \(\theta\) into the equation for \(\frac{d\theta}{dt}\):
\[\frac{d\theta}{dt} = \frac{4}{15 \cdot \sec^2(45^\circ)}\]
Since \(\sec(\theta) = \frac{1}{\cos(\theta)}\) and \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\):
\[\frac{d\theta}{dt} = \frac{4}{15 \cdot (\frac{2}{\sqrt{2}})^2}\]
\[\frac{d\theta}{dt} = \frac{4}{15 \cdot \frac{4}{2}}\]
\[\frac{d\theta}{dt} = \frac{4}{15 \cdot 2}\]
\[\frac{d\theta}{dt} = \frac{1}{15}\]
Therefore, the rate at which the angle of inclination of the observer's line of sight is increasing when the balloon is 300m above the ground is \(\frac{1}{15}\) radians per minute.