help me solve this | 6 parts remaining\nan observer stands 285 ft from the launch site of a hot - air…

help me solve this | 6 parts remaining\nan observer stands 285 ft from the launch site of a hot - air balloon at an elevation equal to the elevation of the launch site. the balloon is launched vertically and maintains a constant upward velocity of 30 ft/s. what is the rate of change of the angle of elevation of the balloon when it is 380 ft from the ground? (hint: the angle of elevation is the angle θ between the observers line of sight to the balloon and the ground.)\nthe given rate is the rate of change in the height of the balloon, or \\( \\frac{dy}{dt}=30 \\) ft/s. the goal is to find the rate \\( \\frac{d\\theta}{dt} \\) when the balloon is \\( y = 380 \\) ft above the launch site.\nfirst, write an equation that relates the angle θ to the height of the balloon y\n

help me solve this | 6 parts remaining\nan observer stands 285 ft from the launch site of a hot - air balloon at an elevation equal to the elevation of the launch site. the balloon is launched vertically and maintains a constant upward velocity of 30 ft/s. what is the rate of change of the angle of elevation of the balloon when it is 380 ft from the ground? (hint: the angle of elevation is the angle θ between the observers line of sight to the balloon and the ground.)\nthe given rate is the rate of change in the height of the balloon, or \\( \\frac{dy}{dt}=30 \\) ft/s. the goal is to find the rate \\( \\frac{d\\theta}{dt} \\) when the balloon is \\( y = 380 \\) ft above the launch site.\nfirst, write an equation that relates the angle θ to the height of the balloon y\n

Answer

Explanation:

Step1: Write the trigonometric relationship

We know that in a right - triangle, (\tan\theta=\frac{y}{285}) (where (y) is the height of the balloon and (285) is the horizontal distance from the observer to the launch site).

Step2: Differentiate both sides with respect to time (t)

Using the chain rule, (\frac{d}{dt}(\tan\theta)=\frac{d}{dt}(\frac{y}{285})). The derivative of (\tan\theta) with respect to (t) is (\sec^{2}\theta\frac{d\theta}{dt}), and the derivative of (\frac{y}{285}) with respect to (t) is (\frac{1}{285}\frac{dy}{dt}). So we have (\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{285}\frac{dy}{dt}).

Step3: Find (\sec\theta) when (y = 380)

When (y = 380), using the Pythagorean theorem in the right - triangle (hypotenuse (r=\sqrt{285^{2}+380^{2}})), and (\sec\theta=\frac{r}{285}). Also, (\tan\theta=\frac{380}{285}=\frac{4}{3}), and (\sec^{2}\theta=1 + \tan^{2}\theta). So (\sec^{2}\theta=1+\left(\frac{4}{3}\right)^{2}=1+\frac{16}{9}=\frac{9 + 16}{9}=\frac{25}{9}).

Step4: Solve for (\frac{d\theta}{dt})

We know that (\frac{dy}{dt}=30) ft/s. Substituting (\sec^{2}\theta=\frac{25}{9}) and (\frac{dy}{dt}=30) into (\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{285}\frac{dy}{dt}), we get (\frac{25}{9}\frac{d\theta}{dt}=\frac{30}{285}). Then (\frac{d\theta}{dt}=\frac{30\times9}{285\times25}=\frac{270}{7125}=\frac{18}{475}\approx0.038) rad/s.

Answer:

(\frac{18}{475}\text{ rad/s}\approx0.038\text{ rad/s})