Question

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Answer 1

We are asked to determine the value of the variation of the angle with respect to "x". To do that we will determine an equation that relates "x" and the angle using the following right triangle:

We can use the function tangent since this is defined as:

$\tan\theta=\frac{opposite}{adjacent}$

Now, we plug in the values:

$\tan\theta=\frac{3000}{x}$

Now, we will determine the angle when "x = 2000". We substitute 2000 in the equation:

$\tan\theta=\frac{3000}{2000}$

Now, we take the inverse function of the tangent:

$\theta=\tan^{-1}(\frac{3000}{2000})$

Solving the operations we get:

$\theta=56.31°$

Therefore, the angle is 56.31 degrees or 0.983 radians.

We are asked to determine the value of the derivative of "x" with respect to "t". Since this is equivalent to the velocity of the plane we have that:

$\frac{dx}{dt}=-200\frac{ft}{s}$

Now, we are asked to determine the derivative of the angle with respect to time. To do that we will determine the derivative with respect to time on both sides of the equation:

$\frac{d}{dt}(\tan\theta)=\frac{d}{dt}(\frac{3000}{x})$

For the derivative on the left side, we know that the derivative of the tangent is the secant squared, therefore, we have:

$\sec^2(\theta)\frac{d\theta}{dt}=\frac{d}{dt}(\frac{3000}{x})$

Now, we determine the derivative on the right side using the following formula:

$\frac{d}{dx}(\frac{a}{x})=-\frac{a}{x^2}$

Applying the rule we get:

$\sec^2\theta\frac{d\theta}{dt}=-\frac{3000}{x^2}\frac{dx}{dt}$

Now, we solve for the derivative of the angle with respect to time:

$\frac{d\theta}{dt}=-\frac{1}{\sec^2\theta}(\frac{3000}{x^2})\frac{dx}{dt}$

Now, we substitute the values:

$\frac{d\theta}{dt}=-\frac{1}{\sec^2(0.983)}(\frac{3000ft}{(2000ft)^2})(-200\frac{ft}{s})$

Solving the operations:

$\frac{d\theta}{dt}=0.047\frac{rad}{s}$

Therefore, the variation of the angle with respect to time is 0.047 rad/s.

This means that the dish is rotating at a speed of 0.047 rad/s.