Question

The first thing I did a lot of people who

Answer 1

Hello there. To solve this question, we have to remember some properties about derivatives.

Given the following expression:

$\begin{gathered} \sin(\theta)=\dfrac{x}{z} \\ \end{gathered}$

Whereas x, z and θ are functions of the time, t.

We know that

$\dfrac{\mathrm{d}x}{\mathrm{d}t}=-60$

We want to determine the value for

$\dfrac{\mathrm{d}\theta}{\mathrm{d}t}$

When z = 2, θ = π/6 and

$z=2\text{ miles},\text{ }\theta=\dfrac{\pi}{6}\text{ radians and }\dfrac{\mathrm{d}z}{\mathrm{d}t}=-55\text{ mph}$

For this, we have to use implicit differentiation.

Since the variables are function of time, we differentiate both sides of the equation with respect to the time t.

$\dfrac{\mathrm{d}}{\mathrm{d}t}(\sin(\theta))=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{x}{z}\right)$

Knowing the Chain Rule:

$\dfrac{d}{dt}(f(x(t))=\dfrac{d(f(x(t))}{dx}\cdot\dfrac{dx}{dt}$

And the derivative of the sine and quotient rules, we get

$\cos(\theta)\cdot\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=\dfrac{\dfrac{\mathrm{d}x}{\mathrm{d}t}\cdot\,z-x\cdot\dfrac{\mathrm{d}z}{\mathrm{d}t}}{{z^2}}$

Plugging the values, we get

$\cos\left(\dfrac{\pi}{6}\right)\cdot\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=\dfrac{(-60)\cdot2-x\cdot(-55)}{2^2}$$\cos\left(\dfrac{\pi}{6}\right)\cdot\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=\dfrac{(-60)\cdot2-x\cdot(-55)}{2^2}$

Of course, to find the value of x, we use the values for the angle and z in the former expression:

$\sin\left(\dfrac{\pi}{6}\right)=\dfrac{x}{2}\Rightarrow\dfrac{1}{2}=\dfrac{x}{2}\Rightarrow x=1$

Therefore we get

$\begin{gathered} \dfrac{\sqrt{3}}{2}\cdot\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=\dfrac{-120+55}{4} \\ \\ \dfrac{\sqrt{3}}{2}\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=-\dfrac{65}{4} \\ \\ \dfrac{\mathrm{d}\theta}{\mathrm{d}t}=-\dfrac{65}{2\sqrt{3}} \end{gathered}$

Using a calculator, we find a number accurate to two decimal places:

$\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\approx-18.76\text{ radians per hour}$

or rph.