My question is similar to Avik's but with different values.
A camera, located 4 km from the launch pad, is tracking the rocket that is fired straight up. When the height of the rocket is 13 km, the camera is rotating at the rate of 3/169 radians per second. What is the speed of the rocket at that instant? Give your answer in km/sec
I do not know how to figure out the value for sec^2 theta.
Thanks for all the help.
answered above.
To find the speed of the rocket at that instant, we need to find the derivative of the height function with respect to time. In this case, the height function is given as h(t) = 13 km.
However, we are also given the rate at which the camera is rotating, which is given as dθ/dt = 3/169 radians per second. We can relate the angle between the camera and the rocket to the height of the rocket using trigonometry.
Let's assume that θ represents the angle between the camera and the rocket. We can construct a right triangle with the base representing the horizontal distance between the camera and the launch pad (4 km), and the height representing the height of the rocket above the launch pad (13 km).
Now, we can use trigonometry to relate the angle θ to the height h. Since the tangent function relates the opposite side to the adjacent side of a right triangle, we have:
tan(θ) = h / 4
Next, we can differentiate both sides of this equation with respect to time, using the chain rule:
sec^2(θ) * dθ/dt = dh/dt / 4
We are given that dθ/dt = 3/169 radians per second, and we need to find sec^2(θ). To solve for this value, we can rearrange the equation to isolate sec^2(θ):
sec^2(θ) = (1/4) * (dh/dt) / (dθ/dt)
Now we can substitute the given values into this equation. We know that dh/dt = 0 since the height of the rocket is constant at 13 km. We also know that dθ/dt = 3/169 radians per second. Plugging in these values, we get:
sec^2(θ) = (1/4) * (0) / (3/169) = 0
Therefore, sec^2(θ) is equal to 0.
Finally, to find the speed of the rocket at that instant, we need to find the derivative of the height function with respect to time. Since the height of the rocket is constant at 13 km, the derivative is zero. The speed of the rocket at that instant is therefore 0 km/sec.
In summary, by using trigonometry and differentiation, we were able to find that the speed of the rocket at that instant is 0 km/sec.