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The first thing I did a lot of people who

The first thing I did a lot of people who-example-1
User Kevin Lamb
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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.

User Spodger
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