Problem 31
Question
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?
Step-by-Step Solution
Verified Answer
The maximum angle of descent during the landing approach is approximately \(0.43^\circ\).
1Step 1: 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:
\(f'(x) = 1.30212\times 10^{-9}x^2 - 3.125\times 10^{-5}x\)
2Step 2: 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 \(x\):
\(1.30212\times 10^{-9}x^2 - 3.125\times 10^{-5}x = 0\)
3Step 3: 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:
\(x = 0\)
\(1.30212\times 10^{-9}x - 3.125\times 10^{-5} = 0\)
4Step 4: 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 \(x = 24000\).
5Step 5: 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 \(0.43^\circ\).
Key Concepts
DifferentiationCritical PointsAngle of Descent
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:
- \(f(x) = 4.3404 \times 10^{-10} x^3 - 1.5625 \times 10^{-5} x^2 + 3000\)
- \(f'(x) = 1.30212 \times 10^{-9} x^2 - 3.125 \times 10^{-5} x\)
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:
- \(1.30212 \times 10^{-9} x^2 - 3.125 \times 10^{-5} x = 0 \)
- \(x (1.30212 \times 10^{-9} x - 3.125 \times 10^{-5}) = 0\)
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:
- \(f'(24000) = -0.0075\)
- \(\theta = \arctan(-0.0075)\)
- \(\theta \approx -0.43^\circ\)
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