Let's draw the scenario to understand it better:
From the figure, the information given are:
[tex]\frac{dy}{d\text{t}}=\text{ 300 ft./sec}[/tex]Question:
[tex]\text{What is }\frac{dD}{dt}\text{ at y = }300\text{ ft.}[/tex]Step 1: We write a function that relates the quantities in the diagram using Pythagorean
Theorem.
[tex]\text{ 400}^2\text{ + }y^2=D^2[/tex]Step 2: Differentiate with respect to t.
[tex]2y\frac{dy}{dt}=\text{ }2D\frac{dD}{dt}[/tex]Now we wish to plug in specific numbers for every quantity in the above equation except for dD/dt. However, we notice that we don’t have a specific value for D at y = 400 ft. So first we need to find D at y = 400 ft. using the Pythagorean Theorem.
[tex]400^2\text{ + }300^2=D^2[/tex][tex]D\text{ = }\sqrt[]{400^2+300^2}[/tex][tex]D\text{ = 500 ft.}[/tex]Step 3: Finish the problem by plugging in numbers for every quantity in the equation
containing dD/dt.
[tex]2(300)(300)\text{ = 2(500)}(\frac{dD}{dt})[/tex][tex]\frac{dD}{dt}=\text{ }\frac{2(300)(300)}{2(500)}[/tex][tex]\frac{dD}{dt}=\text{ }\frac{180,000}{1000}[/tex][tex]\frac{dD}{dt}=180[/tex]Conclusion: When the rocket is 300 ft. feet from the ground, the distance between the observer and the rocket is increasing at a rate of 180 ft./sec.
