Rates of Change — Question 10

Question 10 - Solution

Let:

• $x(t)$: horizontal distance of bicycle from point below the balloon — $\frac{dx}{dt} = 15$

• $y(t)$: vertical distance of balloon from ground — $\frac{dy}{dt} = -2$

• $z(t)$: distance between balloon and bicycle

Using the Pythagorean Theorem: $$z^2 = x^2 + y^2$$

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

Plug in known values: $$x = 60, \quad y = 100, \quad z = \sqrt{60^2 + 100^2} = \sqrt{3600 + 10000} = \sqrt{13600} = 116.6$$

$$z \frac{dz}{dt} = 60(15) + 100(-2) = 900 - 200 = 700 \Rightarrow \frac{dz}{dt} = \frac{700}{116.6} \approx 6.01 \, \text{ft/s}$$

Answer: $$\boxed{\frac{dz}{dt} \approx 6.01 \, \text{ft/s}}$$