Question

A woman standing on a cliff is watching a motorboat through a telescope as the boat approaches the shoreline directly below her. If the telescope is 25 meters above the water level and if the boat is approaching the cliff at 20 m/s, at what rate is the acute angle made by the telescope with the vertical changing when the boat is 250 meters from the shore? 

with a full solution please and please include the given.

Answer 1

The picture is a geometric depiction of the given information.

We introduce variables $x$ and $\theta$ as shown.

We are given that $dx/dt=-20 \ m/s.$ The sign is negative because $x$ is decreasing with time.

From right triangle

$$\tan \theta=\dfrac{x}{25}$$

Differentiate both sides with respect to $t$

$$\dfrac{d}{dt}\big(\tan \theta\big)=\dfrac{d}{dt}\big(\dfrac{x}{25}\big)$$

Use the Chain Rule

$$\dfrac{1}{\cos ^2\theta }\cdot\dfrac{d \theta}{dt}=\dfrac{1}{25}\cdot\dfrac{dx}{dt}$$

$$(1+\tan ^2 \theta)\cdot\dfrac{d \theta}{dt}=\dfrac{1}{25}\cdot\dfrac{dx}{dt}$$

Substitute

$$\bigg(1+\big(\dfrac{x}{25}\big)^2\bigg)\cdot\dfrac{d \theta}{dt}=\dfrac{1}{25}\cdot\dfrac{dx}{dt}$$

Solve for $\dfrac{d \theta}{dt}$

$$(x^2+625)\cdot\dfrac{d \theta}{dt}=25\cdot\dfrac{dx}{dt}$$

$$\dfrac{d \theta}{dt}=\dfrac{25}{x^2+625}\cdot\dfrac{d x}{dt}$$

When $x=250\ m$

$$\dfrac{d \theta}{dt}=\dfrac{25}{(250)^2+625}\cdot(-20\ m/s)$$

$$\dfrac{d \theta}{dt}=-\dfrac{4}{505} \ rad/s$$

$$\dfrac{d \theta}{dt}=-0.008 \ rad/s$$