Someone had helped me yesterday, but I need it broken down more because I honestly don't understand any of this?
A plane flying horizontally at an altitude of 2 mi and a speed of 420 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station.
Using the Pythagorean Theorem from your Algebra I days, the distance to the plane, z, after the plane has gone x miles since being directly overhead, is
z^2 = 2^2 + x^2
we want the rate at which the distance is changing, dz/dt. So, taking derivatives, and remembering the chain rule,
2z dz/dt = 0 + 2x dx/dt
We know that when x=5,
z^2 = 2^2+5^2 = 29, so
z = √29
So, when x=5, we have, using dx/dt=420 mi/hr
2√29 dz/dt = 2*5*420
dz/dt = 4200/(2√29) = 389.96 mi/hr
Thank you very much! I was thinking this was a ratio problem for some reason.
To solve this problem, we'll need to apply the concepts of related rates. Related rates problems involve finding the rate at which one quantity changes with respect to another quantity. In this case, we need to find the rate at which the distance from the plane to the radar station is increasing (let's call it the rate of change of distance) when the plane is 5 miles away from the station.
Let's break down the problem step by step:
1. Identify the given information:
- Altitude of the plane: 2 mi
- Speed of the plane: 420 mi/h
2. Determine the relevant variables:
- Distance from the plane to the radar station: We can represent this distance as "x."
3. Draw a diagram:
Try visualizing the situation in your mind or sketch it on paper. Draw a right triangle with the radar station at one vertex, the plane directly above it, and the distance between them labeled as "x."
4. Establish an equation relating the variables:
By using the Pythagorean theorem, we know that the square of the hypotenuse (distance) is equal to the sum of the squares of the other two sides (altitude and x):
x^2 = (altitude)^2 + (distance)^2
x^2 = 2^2 + (distance)^2
x^2 = 4 + (distance)^2
5. Differentiate both sides of the equation with respect to time (since we are interested in the rate of change):
d(x^2)/dt = d(4 + (distance)^2)/dt
6. Apply the chain rule (differentiating both sides with respect to time):
2x(dx/dt) = 2(distance)(d(distance)/dt)
7. Substitute known values and solve for dx/dt:
We are given that the plane is 5 miles away from the station when we want to find the rate of change of distance.
Plug in x = 5 into the equation and rearrange to solve for dx/dt:
2(5)(dx/dt) = 2(distance)(d(distance)/dt)
10(dx/dt) = 2(distance)(d(distance)/dt)
8. Find the value of "distance" (d(distance)/dt) when x = 5:
To find the value of d(distance)/dt, we need another piece of information. Without it, we cannot solve the problem. If no additional information is provided, assume all variables are changing with respect to time. In this case, let's assume the distance between the plane and the radar station is changing at a constant rate. Therefore, d(distance)/dt = dx/dt.
9. Substitute dx/dt = d(distance)/dt into the equation:
10(dx/dt) = 2(distance)(d(distance)/dt)
10(dx/dt) = 2(distance)(dx/dt)
10. Solve for dx/dt:
Divide both sides of the equation by 2(distance):
10(dx/dt) / (2(distance)) = dx/dt
5(dx/dt) / distance = dx/dt
11. Simplify the equation and find the rate of change of distance:
Since dx/dt is a common term, it cancels out on both sides of the equation:
5 / distance = 1/(dx/dt)
dx/dt = 1 / (5 / distance)
dx/dt = distance / 5
12. Plug in the given value of distance (5 miles) into the equation:
dx/dt = 5 / 5
dx/dt = 1
Therefore, the rate at which the distance from the plane to the radar station is increasing when it is 5 miles away from the station is 1 mile per hour.