An airplane flies at an altitude of 5mies toward a point directly over an observer. The speed of the plain is 600 miles per hour. Find the rate at which the angle of elevation thetha is changing when the angle is 75 degrees.
Please help with this question.
Plain or plane
theta or thetha
miles or mies
What angle? Is it the angle of elevation?
Then if angle of elevation, vertical is 5miles, horizontal is 5cos75
horizontal= 5cosTheta
dh/dt= -5sin Theta dTheta/dt
600=-5sin75 dtheta/dt
solve for dTheta/dt
Plain
θ
miles
To find the rate at which the angle of elevation, theta (θ), is changing, we can use trigonometry and calculus. Let's break down the problem and explain each step:
1. Draw a diagram: Draw a right triangle with the observer at the base and the airplane flying above it. Label the base of the triangle as "x" (the horizontal distance) and the altitude of the airplane as "y" (the vertical distance).
2. Find the relationship between x, y, and theta: Using trigonometry, we can see that the tangent of theta is equal to y divided by x. We have tan(θ) = y / x.
3. Differentiate the equation: To find the rate at which theta is changing with respect to time, we need to differentiate the equation in terms of time. We differentiate both sides of the equation with respect to t (time) using implicit differentiation.
d(tan(θ))/dt = d(y/x)/dt
4. Substitute the given values: We are given that the airplane flies at an altitude of 5 miles, which we can convert into feet (5 miles * 5280 feet/mile). We also know that the speed of the plane is 600 miles per hour.
5. Calculate the values of x and y: Let's determine the values of x and y using the given information. We can use the Pythagorean theorem to find the relationship between x, y, and the altitude (5 miles).
x^2 + y^2 = (5 miles)^2
6. Solve for x: Since x is the horizontal distance, we can solve for x using the equation x^2 + y^2 = (5 miles)^2. Rearrange the equation:
x^2 = (5 miles)^2 - y^2
x = sqrt((5 miles)^2 - y^2)
7. Substitute x into the equation: Replace x with sqrt((5 miles)^2 - y^2) in the equation tan(θ) = y / x.
tan(θ) = y / sqrt((5 miles)^2 - y^2)
8. Differentiate with respect to time: Differentiate both sides of the equation with respect to t (time).
d(tan(θ))/dt = d(y / sqrt((5 miles)^2 - y^2))/dt
9. Solve for d(θ)/dt: Since we want to find the rate at which θ is changing, we are looking for d(θ)/dt.
d(θ)/dt = d(tan(θ))/dt / d(y / sqrt((5 miles)^2 - y^2))/dt
10. Use the chain rule: Using the chain rule, we can find the derivatives of both sides of the equation.
d(θ)/dt = (sec^2(θ)) * (dθ/dt) / ((d(sqrt((5 miles)^2 - y^2)))/dy) * (dy/dt)
11. Evaluate the derivatives: We can evaluate the derivatives based on the given information. The rate at which the angle is changing (dθ/dt) and the rate at which the altitude is changing (dy/dt) are not provided in the question. You will need to know these rates to calculate the final answer.
By following these steps, you can find the rate at which the angle of elevation, theta, is changing when the angle is 75 degrees. Remember to substitute the given values and solve for the unknown rates (dθ/dt and dy/dt) to get the final numerical answer.