Problem 93
Question
The view \(V,\) or distance in miles, that one can see to the horizon from a height \(h,\) in feet, is given by $$ V=1.22 \sqrt{h} $$. a) Find the rate of change of \(V\) with respect to \(h\). b) How far can one see to the horizon from an airplane window at a height of \(40,000 \mathrm{ft} ?\) c) Find the rate of change at \(h=40,000\). d) Explain the meaning of your answer to part (c).
Step-by-Step Solution
Verified Answer
a) \( \frac{1.22}{2\sqrt{h}} \)
b) 244 miles
c) 0.00305 miles/foot
d) It means each foot increases view distance by 0.00305 miles at 40,000 ft.
1Step 1: Calculate the rate of change of V with respect to h
To find the rate of change of \( V \) with respect to \( h \), calculate the derivative of \( V \) with respect to \( h \). Start with the given formula:\[ V = 1.22 \sqrt{h} \]The derivative \( \frac{dV}{dh} \) is obtained by applying the power rule:\[ \frac{dV}{dh} = \frac{1.22}{2} h^{-1/2} = \frac{1.22}{2\sqrt{h}} \]So, the rate of change is \( \frac{1.22}{2\sqrt{h}} \).
2Step 2: Calculate V for h = 40,000 feet
Substitute \( h = 40,000 \) into the formula for \( V \):\[ V = 1.22 \sqrt{40000} \]Calculate \( \sqrt{40000} = 200 \). Therefore:\[ V = 1.22 \times 200 = 244 \]Thus, one can see 244 miles to the horizon from 40,000 feet.
3Step 3: Calculate the rate of change for h = 40,000
Using the derivative found in Step 1, substitute \( h = 40,000 \):\[ \frac{dV}{dh} = \frac{1.22}{2\sqrt{40000}} = \frac{1.22}{2 \times 200} \]Calculate \( \frac{1.22}{400} = 0.00305 \).Thus, the rate of change of \( V \) with respect to \( h \) at \( h = 40,000 \) is approximately 0.00305 miles per foot.
4Step 4: Explain the meaning of the rate of change
In part (c), the calculated rate of change means that for each additional foot in height at 40,000 feet, the distance to the horizon increases by approximately 0.00305 miles. This indicates how the visibility distance grows as you rise above a certain height.
Key Concepts
Rate of ChangeDerivativesPower RuleDistance to Horizon
Rate of Change
The rate of change tells us how one quantity changes in relation to another. In our exercise, we're curious about how the view, or the distance to the horizon, changes with the height from which it is observed. This concept is foundational in calculus as it helps to understand the relationship between different variables.
In practical terms, when we talk about the rate of change here, we're looking at how much extra distance to the horizon you can see as you climb higher. For instance, if you move up just one foot, how much farther will you be able to see? The answer to this is not constant but changes depending on your current height due to the nature of the square root function that defines our distance formula.
The result of the rate of change gives us tangible insights into the behavior of the system we're studying, in this case, our view to the horizon.
In practical terms, when we talk about the rate of change here, we're looking at how much extra distance to the horizon you can see as you climb higher. For instance, if you move up just one foot, how much farther will you be able to see? The answer to this is not constant but changes depending on your current height due to the nature of the square root function that defines our distance formula.
The result of the rate of change gives us tangible insights into the behavior of the system we're studying, in this case, our view to the horizon.
Derivatives
Derivatives are a core concept in calculus, providing a method to calculate the rate of change. By finding the derivative of a function, we can understand how that function's output changes as its input changes. In our problem, we used the derivative of the viewing distance formula, \(V = 1.22 \sqrt{h}\), with respect to height \(h\). This gives us \(\frac{dV}{dh} = \frac{1.22}{2\sqrt{h}}\).
This derivative tells us the instantaneous rate of change of the distance to the horizon as the height changes. It's like peeking at the speedometer, understanding at each moment how quickly your view is extending as you rise.
To find this derivative, we employ some calculus rules like the power rule. It's important to carefully follow these rules to correctly find how the function changes over its range.
This derivative tells us the instantaneous rate of change of the distance to the horizon as the height changes. It's like peeking at the speedometer, understanding at each moment how quickly your view is extending as you rise.
To find this derivative, we employ some calculus rules like the power rule. It's important to carefully follow these rules to correctly find how the function changes over its range.
Power Rule
The power rule is a handy shortcut for finding the derivative of expressions where a variable is raised to a power. The general form of the power rule states that for \(f(x) = x^n\), the derivative is \(f'(x) = n \cdot x^{n-1}\).
In our distance formula, \(V = 1.22 \sqrt{h}\), we can rewrite \(\sqrt{h}\) as \(h^{1/2}\). Applying the power rule here, we differentiate \(h^{1/2}\) to obtain \(\frac{1}{2}h^{-1/2}\).
This rule makes it possible to quickly find how the rate of change behaves without laboriously applying limits or other fundamental methods. It's a powerful tool in the calculus toolkit that simplifies complex-looking expressions into easily manageable forms.
In our distance formula, \(V = 1.22 \sqrt{h}\), we can rewrite \(\sqrt{h}\) as \(h^{1/2}\). Applying the power rule here, we differentiate \(h^{1/2}\) to obtain \(\frac{1}{2}h^{-1/2}\).
This rule makes it possible to quickly find how the rate of change behaves without laboriously applying limits or other fundamental methods. It's a powerful tool in the calculus toolkit that simplifies complex-looking expressions into easily manageable forms.
Distance to Horizon
Distance to the horizon is an interesting concept that combines geometry and calculus. It refers to how far one can see across the Earth's surface from a given height. Due to the Earth's curvature, the distance we are able to see extends as we rise above the surface.
In this exercise, the formula \(V = 1.22 \sqrt{h}\) gives the distance in miles one can see from a height \(h\) in feet. Increasing your height gives a higher vantage point, allowing a wider field of vision to the Earth's edge.
This relationship is crucial, especially in fields such as aviation and meteorology, where understanding visibility can impact navigation and safety. It's amazing to think how a mathematical formula can accurately predict such visible distances based solely on height. Understanding this formula means not only grasping the underlying calculus but also appreciating the real-world applications that come into play when we're discussing views that stretch to the horizon and beyond.
In this exercise, the formula \(V = 1.22 \sqrt{h}\) gives the distance in miles one can see from a height \(h\) in feet. Increasing your height gives a higher vantage point, allowing a wider field of vision to the Earth's edge.
This relationship is crucial, especially in fields such as aviation and meteorology, where understanding visibility can impact navigation and safety. It's amazing to think how a mathematical formula can accurately predict such visible distances based solely on height. Understanding this formula means not only grasping the underlying calculus but also appreciating the real-world applications that come into play when we're discussing views that stretch to the horizon and beyond.
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