Question

A camera is filming the progress as a daredevil attempts to scale the wall of a skyscraper. The climber is moving vertically at a constant rate of 16 feet per minute, and the camera is 400 feet from the base of skyscraper. Through how many radians per minute is the camera angle changing when the climber is 300 feet up the building?

Answer 1

To find the rate at which the camera angle is changing when the climber is 300 feet up the building, we can use trigonometry and take the derivative of the equation.

Step-by-step explanation:

To find the rate at which the camera angle is changing, we can use trigonometry. Let's assume that the angle between the camera's line of sight and the ground is 'θ'. Now, the height at which the climber is located can be represented as 'h'. Using the tangent function, we can write the equation: tan(θ) = h/400.

To find how fast the angle is changing, we need to take the derivative of this equation with respect to time (t):

sec^2(θ) * dθ/dt = (1/400) * dh/dt.

We know that the climber is moving vertically at a constant rate of 16 feet per minute, so dh/dt = 16 feet per minute. Plugging in this value and the given height of 300 feet, we can solve for dθ/dt.

Answer 2

The rate at which the camera angle is changing can be found using trigonometry and differentiation. By setting up a right triangle and using the tangent function, we can relate the rate of change of the camera angle to the rate at which the climber is moving up the building. Substituting the given values into the equation, we can find that the camera angle is changing at a rate of approximately 17.51 radians per minute when the climber is 300 feet up the building.

Step-by-step explanation:

To find the rate at which the camera angle is changing, we can use trigonometry. Since the climber is moving vertically at a constant rate, the camera angle is changing as the climber moves up the building. We can use the tangent function to relate the rate of change of the camera angle to the rate at which the climber is moving up the building. The camera angle is the angle between the line connecting the camera to the climber and the vertical line. We can set up a right triangle where the adjacent side represents the horizontal distance between the camera and the climber, the opposite side represents the height the climber has reached, and the hypotenuse represents the straight line distance from the camera to the climber. Let's call the camera angle theta, the horizontal distance x, and the height y.

Using the tangent function, we have:

tan(theta) = y/x

Now, we can differentiate both sides with respect to time to find the rate of change of the camera angle:

sec^2(theta) * d(theta)/dt = (dy/dt * x - y * dx/dt)/x^2

Since we are given that the climber is moving vertically at a constant rate of 16 feet per minute, we have dy/dt = 16. We are also given that the camera is 400 feet from the base of the skyscraper, so x = 400. And we're asked to find the rate of change of the camera angle when the climber is 300 feet up the building, so y = 300.

Substituting these values into the equation, we have:

sec^2(theta) * d(theta)/dt = (16 * 400 - 300 * dx/dt)/400^2

Now we can solve for d(theta)/dt:

d(theta)/dt = (16 * 400 - 300 * dx/dt)/400^2 * sec^2(theta)

Since we need to find the rate of change of the camera angle when the climber is 300 feet up the building, we can substitute y = 300 into the equation and solve for dx/dt:

tan(theta) = 300/400

theta = arctan(0.75) ≈ 0.6435 radians

Now we can substitute these values into the equation to find dx/dt:

d(theta)/dt = (16 * 400 - 300 * dx/dt)/400^2 * sec^2(0.6435)

Substituting the values we know, we can solve for dx/dt:

d(theta)/dt = (16 * 400 - 300 * dx/dt)/400^2 * sec^2(0.6435)

This gives us dx/dt ≈ 17.51 feet per minute.

Therefore, the camera angle is changing at a rate of approximately 17.51 radians per minute when the climber is 300 feet up the building.