Question

An observer stands 400 ft away from a launch pad to observe a rocket launch. The rocket blasts off and maintains a velocity of 300 ft/sec. Assume the scenario can be modeled as a right triangle. How fast is the observer to rocket distance changing when the rocket is 300 ft from the ground?

Answer 1

Let's draw the scenario to understand it better:

From the figure, the information given are:

`\frac{dy}{d\text{t}}=\text{ 300 ft./sec}`

Question:

`\text{What is }(dD)/(dt)\text{ at y = }300\text{ ft.}`

Step 1: We write a function that relates the quantities in the diagram using Pythagorean

Theorem.

`\text{ 400}^2\text{ + }y^2=D^2`

Step 2: Differentiate with respect to t.

`2y(dy)/(dt)=\text{ }2D(dD)/(dt)`

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.

`400^2\text{ + }300^2=D^2`
`D\text{ = }\sqrt[]{400^2+300^2}`
`D\text{ = 500 ft.}`

Step 3: Finish the problem by plugging in numbers for every quantity in the equation

containing dD/dt.

`2(300)(300)\text{ = 2(500)}((dD)/(dt))`
`(dD)/(dt)=\text{ }(2(300)(300))/(2(500))`
`(dD)/(dt)=\text{ }(180,000)/(1000)`
`(dD)/(dt)=180`

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.