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)=(x)/(z) \\ \end{gathered}`

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

We know that

`\frac{\mathrm{d}x}{\mathrm{d}t}=-60`

We want to determine the value for

`\frac{\mathrm{d}\theta}{\mathrm{d}t}`

When z = 2, θ = π/6 and

`z=2\text{ miles},\text{ }\theta=(\pi)/(6)\text{ radians and }\frac{\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.

`\frac{\mathrm{d}}{\mathrm{d}t}(\sin(\theta))=\frac{\mathrm{d}}{\mathrm{d}t}\left((x)/(z)\right)`

Knowing the Chain Rule:

`(d)/(dt)(f(x(t))=(d(f(x(t)))/(dx)\cdot(dx)/(dt)`

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

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

Plugging the values, we get

`\cos\left((\pi)/(6)\right)\cdot\frac{\mathrm{d}\theta}{\mathrm{d}t}=((-60)\cdot2-x\cdot(-55))/(2^2)`
`\cos\left((\pi)/(6)\right)\cdot\frac{\mathrm{d}\theta}{\mathrm{d}t}=((-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((\pi)/(6)\right)=(x)/(2)\Rightarrow(1)/(2)=(x)/(2)\Rightarrow x=1`

Therefore we get

`\begin{gathered} (√(3))/(2)\cdot\frac{\mathrm{d}\theta}{\mathrm{d}t}=(-120+55)/(4) \\ \\ (√(3))/(2)\frac{\mathrm{d}\theta}{\mathrm{d}t}=-(65)/(4) \\ \\ \frac{\mathrm{d}\theta}{\mathrm{d}t}=-(65)/(2√(3)) \end{gathered}`

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

`\frac{\mathrm{d}\theta}{\mathrm{d}t}\approx-18.76\text{ radians per hour}`

or rph.