Rates of Change — Question 9

Question 9 - Solution

Let:

• $x(t)$: distance from car to balloon launch point (horizontal) — $\frac{dx}{dt} = 60$

• $y(t)$: height of balloon — $\frac{dy}{dt} = 5$

• $z(t)$: distance between car and balloon (hypotenuse)

We apply the Pythagorean theorem: $$z^2 = x^2 + y^2$$

Differentiate both sides: $$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \Rightarrow z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}$$

Plug in values: $$x = 80, \quad y = 150 \Rightarrow z = \sqrt{80^2 + 150^2} = \sqrt{6400 + 22500} = \sqrt{28900} = 170$$

Now plug in: $$170 \cdot \frac{dz}{dt} = 80 \cdot 60 + 150 \cdot 5 \Rightarrow 170 \cdot \frac{dz}{dt} = 4800 + 750 = 5550 \Rightarrow \frac{dz}{dt} = \frac{5550}{170} = 32.65 \, \text{ft/s}$$

Answer: $$\boxed{\frac{dz}{dt} = 32.65 \, \text{ft/s}}$$