/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 The path of an airplane on its f... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 31

The path of an airplane on its final approach to landing is described by the equation \(y=f(x)\) with \(f(x)=4.3404 \times 10^{-10} x^{3}-1.5625 \times 10^{-5} x^{2}+3000\) \(0 \leq x \leq 24,000\) where \(x\) and \(y\) are both measured in feet. a. Plot the graph of \(f\) using the viewing window \([0,24000] \times[0,3000] .\) b. Find the maximum angle of descent during the landing approach. Hint: When is \(d y / d x\) smallest?

Short Answer

Expert verified
The maximum angle of descent during the landing approach is approximately \(0.43^\circ\).

Step by step solution

01

1. Find the derivative of the function \(f(x)\)

Firstly, to find the maximum angle of descent, we will have to find the derivative of \(f(x)\) with respect to \(x\). The function is given as: \(f(x) = 4.3404\times 10^{-10}x^3 - 1.5625\times 10^{-5}x^2 + 3000\) Differentiate \(f(x)\) with respect to \(x\): \(f'(x) = \frac{d}{dx}(4.3404\times 10^{-10}x^3 - 1.5625\times 10^{-5}x^2 + 3000)\) Applying the Power Rule to each term: \(f'(x) = 3 \times 4.3404\times 10^{-10}x^2 - 2 \times 1.5625\times 10^{-5}x\) Simplify the expression: \(f'(x) = 1.30212\times 10^{-9}x^2 - 3.125\times 10^{-5}x\)
02

2. Find the critical points of the derivative

To find the critical points when the angle of descent is maximum, we should set the derivative \(f'(x)\) equal to zero and solve for \(x\): \(1.30212\times 10^{-9}x^2 - 3.125\times 10^{-5}x = 0\)
03

3. Factor the derivative

In this case, the quadratic expression can be factored by taking out the common factors: \(x(1.30212\times 10^{-9}x - 3.125\times 10^{-5}) = 0\) This yields two possible solutions: \(x = 0\) \(1.30212\times 10^{-9}x - 3.125\times 10^{-5} = 0\)
04

4. Solve for \(x\)

Solve the second equation for \(x\): \(1.30212\times 10^{-9}x = 3.125\times 10^{-5}\) \(x = \frac{3.125\times 10^{-5}}{1.30212\times 10^{-9}}\) \(x = 24000\) Thus, the critical points are \(x = 0\) and \(x = 24000\). However, since we are looking for the maximum angle of descent during the approach, we are only interested in the point where the airplane is in the air, i.e., \(x \ne 0\). Therefore, the maximum angle of descent occurs when \(x = 24000\).
05

5. Find the angle of descent when \(x=24000\)

The angle of descent is given by the negative tangent of the rate of change of the height with respect to the horizontal distance. To find the angle of descent in degrees when \(x=24000\), first find the value of \(f'(24000)\): \(f'(24000) = 1.30212\times 10^{-9}(24000)^2 - 3.125\times 10^{-5}(24000)\) \(f'(24000)=- 0.0075\) Now, find the angle in degrees using the arctangent function: \(\theta = \arctan(-0.0075)\) \(\theta \approx -0.43^\circ\) So, the maximum angle of descent during the landing approach is approximately \(0.43^\circ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus that helps us understand how functions change. It involves calculating the derivative, which represents the rate of change of a function with respect to its variable. In our example, the function given is a polynomial describing the flight path of an airplane:
  • \(f(x) = 4.3404 \times 10^{-10} x^3 - 1.5625 \times 10^{-5} x^2 + 3000\)
To find the derivative, \(f'(x)\), we apply the power rule to each term, which gives us:
  • \(f'(x) = 1.30212 \times 10^{-9} x^2 - 3.125 \times 10^{-5} x\)
Differentiation allows us to discover important characteristics of the function, like where it increases or decreases. This is crucial for determining significant features like the angle of descent.
Critical Points
Critical points are where the derivative of a function is zero or undefined. These points can indicate potential maximum or minimum values. For the airplane path, we set the derivative equal to zero:
  • \(1.30212 \times 10^{-9} x^2 - 3.125 \times 10^{-5} x = 0 \)
By factoring, we find:
  • \(x (1.30212 \times 10^{-9} x - 3.125 \times 10^{-5}) = 0\)
This gives possible solutions of \(x = 0\) and \(x = 24000\). Since we are interested in when the airplane is airborne, the relevant critical point is \(x = 24000\). Understanding critical points helps identify where the slope changes direction, leading to maximum angles of descent.
Angle of Descent
The angle of descent is critical for understanding how steeply an airplane descends. It is determined by the negative tangent of the slope of the function at a given point. For \(x=24000\), the derivative is:
  • \(f'(24000) = -0.0075\)
Using the arctangent function, the angle in degrees is:
  • \(\theta = \arctan(-0.0075)\)
  • \(\theta \approx -0.43^\circ\)
Although the angle is negative, it indicates the direction of descent. Calculating this angle helps pilots understand how steeply they are descending during the landing approach. This precise understanding ensures smooth and safe landings. By analyzing the angle of descent, we can evaluate the effectiveness and safety of the landing path.

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Most popular questions from this chapter

Middle-Distance Race As they round the corner into the final (straight) stretch of the bell lap of a middle-distance race, the positions of the two leaders of the pack, \(A\) and \(B\), are given by $$s_{A}(t)=0.063 t^{2}+23 t+15 \quad t \geq 0$$ and $$s_{B}(t)=0.298 t^{2}+24 t \quad t \geq 0$$ respectively, where the reference point (origin) is taken to be the point located 300 feet from the finish line and \(s\) is measured in feet and \(t\) in seconds. It is known that one of the two runners, \(A\) and \(B\), was the winner of the race and the other was the runner- up. a. Show that \(B\) won the race. b. At what point from the finish line did \(B\) overtake \(A\) ? c. By what distance \(\operatorname{did} B\) beat \(A\) ? d. What was the speed of each runner as he crossed the finish line?

Defermine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(f\) is differentiable and \(h(t)=f(a+b t)+f(a-b t)\), then \(h^{\prime}(t)=b f^{\prime}(a+b t)-b f^{\prime}(a-b t) .\)

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