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Chapter 3: Problem 29

If \(\mathrm{H}(\theta)=\theta \sin \theta,\) find \(\mathrm{H}^{\prime}(\theta)\) and \(\mathrm{H}^{\prime \prime}(\theta)\)

Short Answer

Expert verified
\(\mathrm{H}'(\theta) = \sin \theta + \theta \cos \theta\); \(\mathrm{H}''(\theta) = 2 \cos \theta - \theta \sin \theta\).

Step by step solution

01

Recognize the Function

The function given is \(\mathrm{H}(\theta)=\theta \sin \theta\). This is a product of two functions of \(\theta\).
02

Differentiate the Function (First Derivative)

To find \(\mathrm{H}'(\theta)\), we apply the product rule which states if \(u(\theta)=\theta\) and \(v(\theta)=\sin \theta\), then \(\frac{d}{d\theta}(uv) = u'v + uv'\).Here, \(u'(\theta) = 1\) and \(v'(\theta) = \cos \theta\). Applying the product rule, we have:\[\mathrm{H}'(\theta) = 1 \cdot \sin \theta + \theta \cdot \cos \theta = \sin \theta + \theta \cos \theta\]
03

Differentiate Again for the Second Derivative

Now, to find \(\mathrm{H}''(\theta)\), differentiate \(\mathrm{H}'(\theta) = \sin \theta + \theta \cos \theta\) again. This requires using the product rule on \(\theta \cos \theta\) and differentiating \(\sin \theta\) directly.\[\text{Derivative of } \sin \theta = \cos \theta\] \[\text{Using product rule: } \theta \cos \theta \rightarrow 1 \cdot \cos \theta + \theta \cdot (-\sin \theta) = \cos \theta - \theta \sin \theta \] Thus, \[\mathrm{H}''(\theta) = \cos \theta + \cos \theta - \theta \sin \theta = 2 \cos \theta - \theta \sin \theta\]
04

Simplify the Second Derivative

Combine like terms in the expression for \(\mathrm{H}''(\theta)\) to simplify it if possible. In this case, \(\mathrm{H}''(\theta)\) is already simplified to \(2 \cos \theta - \theta \sin \theta\).

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Key Concepts

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

Product Rule
When you have a function that is a product of two differentiable functions, like \ \(H(\theta) = \theta \sin \theta\), the product rule is your best friend. The product rule lets you find the derivative of such multiplied functions easily. It can be remembered with the formula:
  • If \(u(\theta) = \theta\) and \(v(\theta) = \sin \theta\), then \( \frac{d}{d\theta}(uv) = u'v + uv' \).
Here's what this means in simple terms:
  • First, differentiate the first function \( u(\theta) \) while keeping the second function \( v(\theta) \) constant. Multiply them together.
  • Next, keep the first function \( u(\theta) \) constant and differentiate the second function \( v(\theta) \). Multiply them together.
  • Finally, add these two results together to get the derivative of the full product.
This approach simplifies the calculation process by breaking it down into manageable parts.
First Derivative
Finding the first derivative of a function gives us the rate at which it changes. For the function \(H(\theta) = \theta \sin \theta\), applying the product rule helps us identify \(H'(\theta) \).First, we identify the components:
  • \(u'(\theta) = 1\), since the derivative of \(\theta\) is \(1\).
  • \(v'(\theta) = \cos \theta \), as the derivative of \(\sin \theta \) is \(\cos \theta \).
Applying these, the first derivative \(H'(\theta)\) becomes:
  • \(H'(\theta) = 1 \cdot \sin \theta + \theta \cdot \cos \theta\)
  • \(= \sin \theta + \theta \cos \theta\)
Understanding this calculation is crucial as it sets the base for finding further derivatives.
Second Derivative
Looking at the second derivative provides insight into the curvature or concavity of the function. We find this by differentiating the first derivative. For \(H'(\theta) = \sin \theta + \theta \cos \theta\), the steps involve differentiating each term:
  • The derivative of \(\sin \theta\) is \(\cos \theta\).
  • Using the product rule again for \(\theta \cos \theta\), where:
    • \(u'(\theta) = 1 \) and \(v'(\theta) = -\sin \theta\), since \(v(\theta) = \cos \theta\)
    • Resulting in \(\cos \theta - \theta \sin \theta\)
  • Combining, \(H''(\theta) = \cos \theta + \cos \theta - \theta \sin \theta\)
  • \(= 2 \cos \theta - \theta \sin \theta\)
This result, \(2 \cos \theta - \theta \sin \theta\), gives us how sharply the function changes.
Trigonometric Functions
Trigonometric functions like \(\sin\theta\) and \(\cos\theta\) often appear in calculus due to their wave-like properties and periodic nature. They have straightforward derivatives:
  • The derivative of \(\sin \theta\) is \(\cos \theta\).
  • Conversely, the derivative of \(\cos \theta\) is \(-\sin \theta\).
In calculus, these functions can model periodic phenomena - any situation where the behavior repeats in cycles. Understanding these basic derivatives is key when dealing with more complex expressions, including when you're using the product rule on functions containing trigonometric elements. They not only appear in pure math problems but also in physics, engineering, and other real-world applications.

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Most popular questions from this chapter

The rate of change of atmospheric pressure \(\mathrm{P}\) with respect to altitude h is proportional to \(\mathrm{P}\) , provided that the temperature is constant. At \(15^{\circ} \mathrm{C}\) the pressure is 101.3 \(\mathrm{kPa}\) at sea level and 87.14 \(\mathrm{kPa}\) at \(\mathrm{h}=1000 \mathrm{m}\) (a) What is the pressure at an altitude of 3000 \(\mathrm{m}\) ? (b) What is the pressure at the top of Mount McKinley, at an altitude of 6187 \(\mathrm{m}\) ?

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