Question

A hot-air balloon is 150 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 40 mi/hr (58.67 ft/s). if the balloon rises vertically at a rate of 10 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later?

Answer 1

Answer-

Rate of change of the distance between the motorcycle and the balloon after 10 seconds is 57.89 ft/s.

Solution-

Let's consider an instance after t second,

$\text{Distance traveled by the motorcycle in t sec}=\text{Speed}\times \text{t}=58.67t$

$\text{Distance of hot air balloon from ground after t sec}=150+(\text{Speed}\times \text{t})=150+10t$

Let D = Distance between the motorcycle and hot air balloon

As shown in the attachment, the situation represents a right angle triangle, hence applying Pythagoras Theorem,

$\Rightarrow D^2=(58.67t)^2+(150+10t)^2$

$\Rightarrow \dfrac{d}{dt}D^2=\dfrac{d}{dt}(58.67t)^2+\dfrac{d}{dt}(150+10t)^2$

$\Rightarrow 2D\dfrac{dD}{dt}=2\times 58.67^2\times t+2(150+10t)\times 10$

$\Rightarrow D\dfrac{dD}{dt}=58.67^2\times t+(150+10t)\times 10$

$\Rightarrow D\dfrac{dD}{dt}=3442.17t+1500+100t$

$\Rightarrow D\dfrac{dD}{dt}=3542.17t+1500$

$\Rightarrow \dfrac{dD}{dt}=\dfrac{3542.17t+1500}{D}$

After t=10 seconds,

$\Rightarrow D^2=(58.67\times 10)^2+(150+10(10))^2=(586.7)^2+(250)^2=406716.89\\\\\Rightarrow D=\sqrt{406716.89}=637.74$

Then, rate of change of the distance between the motorcycle and the balloon after 10 seconds will be,

$\Rightarrow \dfrac{dD}{dt}=\dfrac{3542.17(10)+1500}{637.74}=57.89\ ft/s$

Answer 2

After 10 seconds, the rate at which the distance increases is 30.91 ft/s.

How to find the rate of change?

Let's define t = 0s as the moment when the motorcycle is below the ballon.

At that moment, we can write the position of the ballon as:

P = (0 ft, 150ft + 10ft/s*t)

And the position of the motorcycle as:

P' = (0ft + 58.67ft/s*t, 0ft).

The distance between the two is given by:

$D = \sqrt{(150ft + 10ft/s*t)^2 + (58.67ft/s*t)^2$

We want to find the rate of change when t = 10s, then we need to differentiate the distance. It will give:

$D' = \frac{2*(10ft/s)^2*t + 2*(10ft/s)*150ft + 2*(58.67ft/s)^2*t}{2*\sqrt{(150ft + 10ft/s*t)^2 + (58.67ft/s*t)^2} }$

Now we need to evaluate that thing on t = 10s.

$D' = \frac{2*(10ft/s)^2*10s + 2*10ft/s*150ft + 2*(58.67ft/s)^2*10s}{2*\sqrt{(150ft + 10ft/s*10s)^2 + (58.67ft/s*10s)^2} } = \frac{39,421.7ft^2/s}{1,275.48ft} \\\\D' = 30.91 ft/s$

So at t = 10s, the rate at which the distance increases is 30.91ft/s.

If you want to learn more about rates of change, you can read:

https://brainly.com/question/8728504