# Calculus 1 relative rates question

• ivanhu1
In summary, the problem involves finding the rates at which the angle of elevation is changing for an airplane flying at a speed of 600 miles per hour and altitude of 8 miles. The equation for the rates is sec^2(θ) * dθ/dt = 8/x^2 * 600, and the sign of dx/dt depends on whether x is increasing or decreasing. The solution involves solving for dθ/dt and evaluating it at θ = 30°, θ = 60°, and θ = 80° in radians.

## Homework Statement

An airplane flies at an altitude of 8 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation θ is changing when the angle is θ = 30°, θ = 60°, and θ = 80°.

## The Attempt at a Solution

I tried solving this problem by my self but i kept getting stuck at one the same spot every time.

(db/dt)=600
y=8

tanθ=8/x

sec^2(θ) * dθ/dt = 8/x^2 * 600

and that's as far i get every time and i don't know if I'm doing it right or not.

ivanhu1 said:

## Homework Statement

An airplane flies at an altitude of 8 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation θ is changing when the angle is θ = 30°, θ = 60°, and θ = 80°.

## The Attempt at a Solution

I tried solving this problem by my self but i kept getting stuck at one the same spot every time.

(db/dt)=600
Makes more sense to call the plane's velocity dx/dt. Also, since the plane is flying toward the observer, x is decreasing, so dx/dt < 0.
ivanhu1 said:
y=8

tanθ=8/x

sec^2(θ) * dθ/dt = 8/x^2 * 600
d/dt(8/x) = d/dt(8x-1) = -8x-2 * dx/dt
Notice that the right side represents a positive number if x > 0.

Other than that, your equation of the rates looks fine. Now solve for dθ/dt, and evaluate it at the three given values of θ. Don't forget to convert the angles to radians.
ivanhu1 said:
and that's as far i get every time and i don't know if I'm doing it right or not.

## 1. What is a relative rate in Calculus 1?

A relative rate in Calculus 1 refers to the change in one quantity in relation to the change in another quantity. It is used to determine how the rate of change of one variable affects the rate of change of another variable.

## 2. How do you find the relative rate in Calculus 1?

To find the relative rate in Calculus 1, you need to take the derivative of the function and then plug in the values of the two variables you want to compare. The resulting number is the relative rate.

## 3. What is the difference between average rate and relative rate in Calculus 1?

The average rate in Calculus 1 is the change in a quantity over a specific interval. It gives a general idea of how the quantity is changing. On the other hand, a relative rate compares the change in one quantity to the change in another quantity, giving a more specific and localized understanding of the relationship between the two variables.

## 4. Can relative rates be negative in Calculus 1?

Yes, relative rates can be negative in Calculus 1. This indicates that as one quantity increases, the other decreases, or vice versa. It is important to pay attention to the signs of relative rates to understand the relationship between the two variables.

## 5. How are relative rates used in real-world applications?

Relative rates are used in real-world applications to analyze and understand the relationships between different variables. For example, in economics, relative rates can be used to determine how changes in one variable, such as price, affect changes in another variable, such as demand. They are also commonly used in physics and engineering to study the relationships between various physical quantities.