Problem 65 - Essential Calculus Early Transcendentals

Question

The motion of a spring that is subject to a frictional force or a damping force (such as a shock absorber in a car) is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a Point on such a spring is $$s(t)=2 e^{-1.5 t} \sin 2 \pi t$$ where \(s\) is measured in centimeters and \(t\) in seconds. Find the velocity after \(t\) seconds and graph both the position and velocity functions for 0\(\leqslant t \leqslant 2\) .

Step-by-Step Solution

Answer

The velocity function is \(v(t) = -3e^{-1.5t} \sin(2\pi t) + 4\pi e^{-1.5t} \cos(2\pi t)\).

Step 1: Understand the Problem

The problem involves finding the velocity function from a given position function of a spring's motion. The position is given by the function \(s(t) = 2e^{-1.5t} \sin(2\pi t)\), and we need to get the velocity \(v(t)\), which is the derivative of \(s(t)\).

Step 2: Differentiate the Position Function

To find the velocity \(v(t)\), take the derivative of the position, \(s(t) = 2e^{-1.5t} \sin(2\pi t)\), with respect to \(t\). We use the product rule: if \(u(t) = 2e^{-1.5t}\) and \(v(t) = \sin(2\pi t)\), then \(v'(t) = u'(t)v(t) + u(t)v'(t)\).

Step 3: Apply the Product Rule

Differentiate \(u(t) = 2e^{-1.5t}\). The derivative is \(u'(t) = -3e^{-1.5t}\) by using the chain rule. Differentiate \(v(t) = \sin(2\pi t)\). The derivative is \(v'(t) = 2\pi \cos(2\pi t)\).

Step 4: Calculate the Derivative

The velocity function is \(v(t) = (-3e^{-1.5t})\sin(2\pi t) + (2e^{-1.5t})(2\pi \cos(2\pi t))\). Simplify to get \(v(t) = -3e^{-1.5t} \sin(2\pi t) + 4\pi e^{-1.5t} \cos(2\pi t)\).

Step 5: Graph Both Functions

Plot the position function \(s(t) = 2e^{-1.5t} \sin(2\pi t)\) and the velocity function \(v(t) = -3e^{-1.5t} \sin(2\pi t) + 4\pi e^{-1.5t} \cos(2\pi t)\) over the interval \(0 \leq t \leq 2\). Use a graphing calculator or software to visualize these functions.

Key Concepts

Exponential Functions

Exponential functions are vital in modeling phenomena that change rapidly. They have the general form \(f(x) = a e^{bx}\), where \(e\) is the base of natural logarithms. The defining characteristic of exponential functions is the constant ratio of change, allowing them to model growth (when \(b > 0\)) or decay (when \(b < 0\)).In our exercise, the function \(s(t) = 2e^{-1.5t} \sin(2\pi t)\) includes an exponential term \(2e^{-1.5t}\), indicating that the motion involves damping over time. This damping is modeled by the negative exponent \(-1.5t\), suggesting the position decreases exponentially.
Exponential decay is common in physical systems, such as the damped motion of a spring, radioactive decay, or cooling of an object. Understanding this concept helps predict how systems evolve over time.

Trigonometric Functions

Trigonometric functions, like sine and cosine, describe oscillatory behavior. They are periodic, meaning they repeat their values in regular intervals. The basic forms are \( \sin x\) and \(\cos x\), with properties that make them excellent for modeling waves and cyclic patterns. In the context of the exercise, \( \sin(2\pi t) \) represents periodic motion. The coefficient \(2\pi\) adjusts the frequency of the sine wave, determining how often the oscillations occur per time unit. Frequency alterations impact how quickly the cycles repeat, crucial for analyzing vibrations or waveforms in physical systems.By understanding trigonometric behavior, we can grasp how components like springs, pendulums, or electrical signals operate with time-based periodic changes.

Product Rule

The product rule is a derivative rule for functions that are products of two or more distinct functions. It states that if you have two functions \(u(t)\) and \(v(t)\), their derivative is given by \( (uv)' = u'v + uv'\), where \(u'\) and \(v'\) represent the derivatives of \(u\) and \(v\) respectively.In our exercise, we use the product rule to differentiate the position function \(s(t) = 2e^{-1.5t} \sin(2\pi t)\). Here, the exponential component \(u(t) = 2e^{-1.5t}\) and trigonometric component \(v(t) = \sin(2\pi t)\) must both be differentiated.

• Derive \(u'(t) = -3e^{-1.5t}\) using the chain rule.
• Derive \(v'(t) = 2\pi \cos(2\pi t)\) by differentiating the sine function.

Solving combinations like these requires skill in applying both the product rule and the chain rule.

Chain Rule

The chain rule is essential for differentiating composite functions. It expresses that the derivative of a composite function \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). This rule is crucial for working with expressions where functions nest inside each other.In our exercise, the chain rule applies when differentiating \(u(t) = 2e^{-1.5t}\). Here, \(e^{-1.5t}\) is a composite function, composed of \(e^x\) and \(-1.5t\). Thus, the derivative \(u'(t)\) results from taking the derivative of the outside function \(e^x\), multiplying by the derivative of the inside function \(-1.5t\), giving us \(-3e^{-1.5t}\).Mastering the chain rule allows seamless manipulation of complex expressions, unraveling layers in functions to expose inner mechanics efficiently.