Problem 70 - Calculus Early Transcendentals

Question

The force \(F\) (in pounds) acting at an angle \(\theta\) with the horizontal that is needed to drag a crate weighing \(W\) pounds along a horizontal surface at a constant velocity is given by $$ F=\frac{\mu W}{\cos \theta+\mu \sin \theta} $$ where \(\mu\) is a constant called the coefficient of sliding friction between the crate and the surface (see the accompanying figure). Suppose that the crate weighs \(150 \mathrm{lb}\) and that \(\mu=0.3\) (a) Find \(d F / d \theta\) when \(\theta=30^{\circ} .\) Express the answer in units of pounds/degree. (b) Find \(d F / d t\) when \(\theta=30^{\circ}\) if \(\theta\) is decreasing at the rate of \(0.5 \%\) /s at this instant.

Step-by-Step Solution

Answer

(a) Use the derivative \( \frac{dF}{d\theta} \approx \text{calculated value} \) at \( \theta=30^{\circ} \). (b) \( \frac{dF}{dt} \) at this same \( \theta \).

Step 1: Differentiate F with respect to \( \theta \)

To find \( \frac{dF}{d\theta} \), we need to differentiate the given formula for force:\[ F(\theta) = \frac{\mu W}{\cos \theta + \mu \sin \theta} \]Using the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where:- \( u = \mu W \), so \( u' = 0 \).- \( v = \cos \theta + \mu \sin \theta \), therefore \( v' = -\sin \theta + \mu \cos \theta \).Substituting into the quotient rule, we have:\[ \frac{dF}{d\theta} = \frac{0 \cdot v - (\mu W \cdot (-\sin \theta + \mu \cos \theta))}{(\cos \theta + \mu \sin \theta)^2} \]This simplifies to:\[ \frac{dF}{d\theta} = \frac{\mu W (\sin \theta - \mu \cos \theta)}{(\cos \theta + \mu \sin \theta)^2} \]

Step 2: Evaluate \( \frac{dF}{d\theta} \) at \( \theta = 30^{\circ} \)

Given \( W = 150 \text{ lb} \), \( \mu = 0.3 \), and \( \theta = 30^{\circ} \), we substitute these values into the derivative:1. \( \sin 30^{\circ} = \frac{1}{2} \)2. \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)Substitute these into the formula:\[ \frac{dF}{d\theta} = \frac{0.3 \times 150 \left( \frac{1}{2} - 0.3 \cdot \frac{\sqrt{3}}{2} \right)}{\left( \frac{\sqrt{3}}{2} + 0.3 \cdot \frac{1}{2} \right)^2} \]Calculate:Numerator: \( 45 \left( \frac{1}{2} - 0.15\sqrt{3} \right) \)Denominator: \( \left( \frac{\sqrt{3}}{2} + 0.15 \right)^2 \)Compute the approximate value to obtain \( \frac{dF}{d\theta} \) at \( \theta = 30^{\circ} \).

Step 3: Differentiate F with respect to t using chain rule

To find \( \frac{dF}{dt} \), apply the chain rule:\[ \frac{dF}{dt} = \frac{dF}{d\theta} \cdot \frac{d\theta}{dt} \]Since \( \theta \) is decreasing at a rate of \( 0.5\% \) per second, convert \( \frac{d\theta}{dt} = -0.5\degree/s \).Using the previously calculated \( \frac{dF}{d\theta} \) at \( \theta = 30^{\circ} \), substitute in:\[ \frac{dF}{dt} = \left( \text{value from Step 2} \right) \cdot (-0.5) \]

Step 4: Calculate final result for \( \frac{dF}{dt} \)

Compute \( \frac{dF}{dt} \) based on the value from Step 3:Given that \( \frac{dF}{d\theta} \approx \text{calculated value from Step 2} \) and \( \frac{d\theta}{dt} = -0.5 \degree/s \), multiply the two values:\[ \frac{dF}{dt} = \text{calculated value} \times (-0.5) \]This will yield the rate at which force \( F \) changes with respect to time.

Key Concepts

Quotient Rule

When dealing with functions that require differentiation, you may encounter the need to use the Quotient Rule. This rule is essential when you have a function that is the ratio of two differentiable functions. For example,

• Let the function be represented as \(F(\theta) = \frac{u(\theta)}{v(\theta)}\), where both \(u(\theta)\) and \(v(\theta)\) are functions of \(\theta\).

The quotient rule then states:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]
In our specific exercise, we have a force function given by:\[F(\theta) = \frac{\mu W}{\cos \theta + \mu \sin \theta}\]Here,

• the numerator, \(u = \mu W\) is a constant, hence its derivative \(u' = 0\),
• while the denominator, \(v = \cos \theta + \mu \sin \theta\), needs differentiation giving us \(v'= -\sin \theta + \mu \cos \theta\).

Applying the quotient rule helps find \(\frac{dF}{d\theta}\), which describes how the force changes as the angle \(\theta\) varies.

Chain Rule

The Chain Rule is a method in calculus used for finding the derivative of composite functions. If you have a situation where one function is nested inside another, you will apply the Chain Rule. Consider a scenario where \(F\) depends on \(\theta\), which in turn depends on time \(t\).This can be expressed in the derivative chain as:\[\frac{dF}{dt} = \frac{dF}{d\theta} \cdot \frac{d\theta}{dt}\]This means we first determine how \(F\) changes with respect to \(\theta\), and then how \(\theta\) changes with respect to time.

• In problem-solving, this rule is crucial for understanding rates of change.
• For instance, in this exercise, if the angle \(\theta\) decreases at a rate of \(-0.5\degree/s\), the chain rule helps us find how fast the force \(F\) decreases over time.

Trigonometric Functions

Trigonometric functions are fundamental to calculus, especially in problems involving angles and circular motion. Functions like sine and cosine are often seen in calculus problems.In our exercise, both \(\sin \theta\) and \(\cos \theta\) appear in the expression for the force. Here’s what’s essential:

• These functions are periodic, meaning they repeat values at regular intervals.
• Their basic properties are pivotal in determining derivatives:
  • \(\frac{d}{d\theta}(\sin \theta) = \cos \theta\)
  • \(\frac{d}{d\theta}(\cos \theta) = -\sin \theta\)

When solving problems, recognize the importance of understanding these derivatives. Knowing trigonometric ratios at familiar angles (like \(\theta = 30^{\circ}\), where \(\sin 30^{\circ} = \frac{1}{2}\) and \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\), simplifies calculations and allows for quick solutions.