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Chapter 3: Problem 23

Find \(d r / d \theta\). $$r=4-\theta^{2} \sin \theta$$

Short Answer

Expert verified
\( \frac{dr}{d\theta} = -2\theta \sin \theta - \theta^{2} \cos \theta \)

Step by step solution

01

Understand the Problem

We're asked to find the derivative \( \frac{dr}{d\theta} \) for the function \( r = 4 - \theta^{2} \sin \theta \). This is essentially finding how the function \( r \) changes with respect to \( \theta \).
02

Differentiate the Constant

The constant term 4 does not depend on \( \theta \), so its derivative with respect to \( \theta \) is 0. Therefore, we can focus on differentiating only \( -\theta^{2} \sin \theta \).
03

Apply the Product Rule

The expression \( -\theta^{2} \sin \theta \) is a product of \( -\theta^{2} \) and \( \sin \theta \). We will use the product rule: if \( u = \theta^{2} \) and \( v = \sin \theta \), then the derivative \( \frac{d}{d\theta} (uv) = u'v + uv' \).
04

Differentiate Each Function Separately

Calculate \( u' = \frac{d}{d\theta}(-\theta^{2}) = -2\theta \) and \( v' = \frac{d}{d\theta}(\sin \theta) = \cos \theta \).
05

Apply the Product Rule

Using the product rule from Step 3, substitute \( u' = -2\theta \), \( u = -\theta^{2} \), \( v = \sin \theta \), and \( v' = \cos \theta \):\[ \frac{d}{d\theta} (-\theta^{2} \sin \theta) = -2\theta \sin \theta + (-\theta^{2} \cos \theta)\]
06

Combine Results

The derivative \( \frac{dr}{d\theta} \) is the sum of the derivatives we've calculated:\[ \frac{dr}{d\theta} = 0 + \left(-2\theta \sin \theta - \theta^{2} \cos \theta\right)\]Recall that the constant term's derivative is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are a fundamental aspect of calculus and mathematics as a whole. These functions relate the angles of a triangle to the lengths of its sides and are vital in understanding periodic phenomena. The two most basic trigonometric functions are sine (\( \sin \theta \)) and cosine (\( \cos \theta \)).
  • **Sine Function:** Represents the y-coordinate of a point on the unit circle and varies smoothly between \(-1\) and \(1\).
  • **Cosine Function:** Represents the x-coordinate on the unit circle, also varying between \(-1\) and \(1\).
These functions play a crucial role in calculus as they often appear in problems involving derivatives and integrals.

In the context of the given problem, we see \( heta^2 \sin \theta \), where \( \sin \theta \) is important in understanding how the function behaves. It's essential to differentiate correctly to see how changes in \( \theta \) affect the overall function.
Product Rule
The product rule is a technique in calculus used when differentiating products of two functions. Given two differentiable functions \( u(\theta) \) and \( v(\theta) \), the derivative of their product \( u(\theta) \cdot v(\theta) \) is derived using:
  • **Formulation of Product Rule:** \[ \frac{d}{d\theta} \left[ u(\theta) \cdot v(\theta) \right] = u'(\theta) \cdot v(\theta) + u(\theta) \cdot v'(\theta) \]
In the original exercise, we applied the product rule to \(-\theta^2 \sin \theta\).Here, the steps include:
  • Set \( u = -\theta^2 \) and find \( u' = -2\theta \)
  • Set \( v = \sin \theta \) and find \( v' = \cos \theta \)
Using the formula, substitute to find the derivative, combining effectively the rates of change. Mastery of this rule is essential in tackling complex calculus problems, particularly those where functions are multiplied together.
Differentiation Techniques
Differentiation techniques are the methods used to compute the derivative of a function. Derivatives help us understand how a function changes, providing insights into the behavior and properties of functions.
  • **Basic Differentiation:** Focuses on differentiating basic functions like powers of \( \theta \), constants, and basic trig functions.
  • **Product Rule:** Discussed previously, crucial for differentiating products of functions.
  • **Chain Rule:** Another key method (though not needed here), for functions composed of other functions.
In solving the problem, basic differentiation allowed us to derive \( u' = -2\theta \) and \( v' = \cos \theta \) individually. We then used these with the product rule to find the derivative of more complex expressions like \(-\theta^2 \sin \theta\). Understanding and mastering these techniques ensures we can handle various functional forms and relationships, making differentiation one of the most powerful tools in calculus.

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Most popular questions from this chapter

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=\sqrt{x} \sin ^{-1} x,[0,1], \quad a=\frac{1}{2}$$

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=\sqrt{3 x-2}, \quad \frac{2}{3} \leq x \leq 4, \quad x_{0}=3$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{2}\left(\frac{x^{2} e^{2}}{2 \sqrt{x+1}}\right)$$

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$T=2 \pi \sqrt{\frac{L}{g}}$$ where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L\). In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2\)

The derivative of \(\cos \left(x^{2}\right)\) Graph \(y=-2 x \sin \left(x^{2}\right)\) for \(-2 \leq\) \(x \leq 3 .\) Then, on the same screen, graph $$y=\frac{\cos \left((x+h)^{2}\right)-\cos \left(x^{2}\right)}{h}$$ for \(h=1.0,0.7,\) and \(0.3 .\) Experiment with other values of \(h\) What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.
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