1. A rocket is fired vertically into the air. Six kilometers away, a telescope tracks the rocket. At a certain moment, the angle between the telescope and the ground is and is increasing at a rate of 0.6 radians per minute. (See the picture. I have defined y to be the height of the rocket in kilometers and θ to be the angle in radians. We’ll also want t to be the number of minutes since launch.) What is the rocket’s velocity at that moment?
a. What derivative is given in the problem? What are its units?
b. What derivative is asked for in the problem? What are its units?
c. Solve the problem
14.4 km/min
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slide 5
Care: they're working with 0.9rad/min
a. The derivative given in the problem is the rate of change of the angle θ with respect to time t, which is stated to be 0.6 radians per minute.
The units of this derivative are radians per minute.
b. The derivative asked for in the problem is the velocity of the rocket at the moment when the angle θ, between the telescope and the ground, is given.
The units of this derivative can be determined by considering the units of the variables involved in the problem. Since the angle θ is measured in radians and is related to the height of the rocket y, which is measured in kilometers, the velocity of the rocket will have units of kilometers per minute.
c. To solve the problem, we need to find the velocity of the rocket at the given moment when the angle θ is known.
The relationship between the angle θ and the height of the rocket y can be described using trigonometric functions. In particular, the tangent function is relevant in this case:
tan(θ) = y / 6
To find the velocity of the rocket, we can differentiate this equation with respect to time t:
d/dt [tan(θ)] = d/dt [y / 6]
Using the chain rule, the left side becomes:
sec^2(θ) * dθ/dt
And the right side becomes:
(1/6) * dy/dt
Since we are given that dθ/dt = 0.6 radians per minute, the equation becomes:
sec^2(θ) * 0.6 = (1/6) * dy/dt
Simplifying the equation, we can solve for dy/dt, which gives us the velocity of the rocket:
dy/dt = (0.6 * 6) / sec^2(θ)
Substituting the given value of the angle θ, we can calculate the rocket's velocity at that moment.