Problem 83
Question
Suppose that a steel ball bearing is released within a vat of fluid and begins to sink. According to one model, the speed \(v(t)\) (in \(\mathrm{m} / \mathrm{s}\) ) of the ball bearing \(t\) seconds after its release is given by the formula $$ v(t)=\frac{9.8}{k}\left(1-e^{-k t}\right) $$ where \(k\) is a positive constant that corresponds to the resistance the fluid offers against the motion of the bearing. (The smaller the value of \(k\), the weaker will be the resistance.) For \(t\) fixed, determine the limiting value of the speed as \(k \rightarrow 0^{+},\) and give a physical interpretation of the limit.
Step-by-Step Solution
Verified Answer
As \(k \to 0^+\), the speed \(v(t)\) approaches \(9.8t\), indicating motion with only gravitational acceleration.
1Step 1: Understanding the Problem
We need to find the limiting value of the speed function \(v(t)\) as the constant \(k\) approaches zero from the positive side. This means we're looking at how the speed behaves as the fluid offers less and less resistance.
2Step 2: Substitute and Simplify the Expression
First, substitute the expression for \(v(t)\) with the given formula: \(v(t) = \frac{9.8}{k}(1-e^{-kt})\). Our goal is to evaluate the limit of this expression as \(k \rightarrow 0^+\).
3Step 3: Analyze the Exponential Term
Focus on the exponential term \(e^{-kt}\). As \(k \rightarrow 0^+\), \(-kt\) approaches 0 for any fixed \(t\), making \(e^{-kt} \rightarrow e^{0} = 1\).
4Step 4: Apply L'Hôpital's Rule
The expression becomes indeterminate of the form \(\frac{0}{0}\), so apply L'Hôpital's Rule by differentiating the numerator and the denominator:\[\lim_{k \to 0^+} \frac{9.8(1-e^{-kt})}{k} = \lim_{k \to 0^+} \frac{9.8(-t e^{-kt})(-1)t}{1} = 9.8t. \]
5Step 5: Conclude the Limit Evaluation
With simplification, the limiting value of \(v(t)\) as \(k\) approaches 0 from the positive side is found to be \(9.8t\).
6Step 6: Interpretation of the Result
Physically, as \(k\) approaches 0, the resistance of the fluid decreases to zero, leading to the bearing accelerating under gravity alone with acceleration \(9.8 \ \mathrm{m/s^2}\), the gravitational acceleration on Earth. Thus, \(v(t) = 9.8t\) represents the speed without any resistance.
Key Concepts
Differential EquationsExponential FunctionsL'Hôpital's Rule
Differential Equations
Differential equations are mathematical tools used to describe the relationship between a function and its derivatives. Imagine them as equations involving unknown functions and their rates of change. In our context, the function in question is the speed of a steel ball bearing sinking through a fluid over time.
By using a differential equation, we can model how different factors—like resistance from the fluid—affect the speed. The given equation describes how the speed changes with time, incorporating both gravitational pulling and the resistance in the fluid.
By using a differential equation, we can model how different factors—like resistance from the fluid—affect the speed. The given equation describes how the speed changes with time, incorporating both gravitational pulling and the resistance in the fluid.
- Functions and their derivatives: Here, the focus is on how speed (our function) changes over time as influenced by the resistance (related to the exponential term).
- Real-world applications: Situations like drop simulations in physics and engineering often utilize differential equations to predict movements and behaviors over time.
Exponential Functions
Exponential functions are mathematical expressions involving exponents. They have the form of an equation with a constant raised to a power, like in our exercise where we see the term \(e^{-kt}\). The exponential function is very useful to model situations where growth or decay is proportional to the current amount, which is typical in scenarios involving resistance or friction.
In the formula \(v(t) = \frac{9.8}{k}(1-e^{-kt})\), the exponential part \(e^{-kt}\) describes how quickly the speed approaches its equilibrium.
In the formula \(v(t) = \frac{9.8}{k}(1-e^{-kt})\), the exponential part \(e^{-kt}\) describes how quickly the speed approaches its equilibrium.
- Decay behavior: The expression \(e^{-kt}\) decreases to 0 as time progresses, highlighting how speed stabilizes when resistance is insignificant.
- Understanding limits: As \(k\) approaches 0, the resistance disappears, meaning the behavior described by the exponential term becomes less significant, allowing the speed to grow linearly with time.
L'Hôpital's Rule
L'Hôpital's Rule is a technique for finding limits of indeterminate forms. It's especially helpful when dealing with expressions that result in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). By differentiating the numerator and the denominator separately, we can simplify the evaluation of a limit.
During our exercise, the limit of the speed function as \(k \to 0^+\) poses an indeterminate form \(\frac{0}{0}\). Using L'Hôpital’s Rule, we efficiently navigated through this challenge.
During our exercise, the limit of the speed function as \(k \to 0^+\) poses an indeterminate form \(\frac{0}{0}\). Using L'Hôpital’s Rule, we efficiently navigated through this challenge.
- Application: Differentiating the terms simplifies expressions, revealing results like reaching \(9.8t\) for our speed as resistance vanishes.
- Practical Importance: For students and professionals alike, being able to evaluate challenging limits using L'Hôpital's Rule can be a powerful tool in calculus and real-world calculations.
Other exercises in this chapter
Problem 81
Find the limit by interpreting the expression as an appropriate derivative. $$\lim _{\Delta x \rightarrow 0} \frac{9\left[\sin ^{-1}\left(\frac{\sqrt{3}}{2}+\De
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Find the limit by interpreting the expression as an appropriate derivative. $$\lim _{w \rightarrow 2} \frac{3 \sec ^{-1} w-\pi}{w-2}$$
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Find the limit by interpreting the expression as an appropriate derivative. $$\lim _{h \rightarrow 0} \frac{\tan ^{-1}(1+h)-\pi / 4}{h}$$
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