/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Find the derivatives of the func... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the derivatives of the functions in Exercises \(19-40\) $$s=\sin \left(\frac{3 \pi t}{2}\right)+\cos \left(\frac{3 \pi t}{2}\right)$$

Short Answer

Expert verified
\(\frac{ds}{dt} = \frac{3 \pi}{2} \left(\cos\left(\frac{3 \pi t}{2}\right) - \sin\left(\frac{3 \pi t}{2}\right)\right)\).

Step by step solution

01

Identify the Function

We need to differentiate the function \(s(t) = \sin\left(\frac{3 \pi t}{2}\right) + \cos\left(\frac{3 \pi t}{2}\right)\). This function is composed of two trigonometric functions: sine and cosine.
02

Differentiate the Sine Component

The derivative of \(\sin(x)\) is \(\cos(x)\). Using the chain rule, the derivative of \(\sin\left(\frac{3 \pi t}{2}\right)\) is \(\frac{d}{dt}\left(\sin\left(\frac{3 \pi t}{2}\right)\right) = \cos\left(\frac{3 \pi t}{2}\right) \cdot \frac{d}{dt}\left(\frac{3 \pi t}{2}\right)\). The derivative of \(\frac{3 \pi t}{2}\) with respect to \(t\) is \(\frac{3 \pi}{2}\). Thus, the derivative of the sine part is \(\cos\left(\frac{3 \pi t}{2}\right) \cdot \frac{3 \pi}{2}\).
03

Differentiate the Cosine Component

The derivative of \(\cos(x)\) is \(-\sin(x)\). Using the chain rule, the derivative of \(\cos\left(\frac{3 \pi t}{2}\right)\) is \(-\sin\left(\frac{3 \pi t}{2}\right) \cdot \frac{d}{dt}\left(\frac{3 \pi t}{2}\right)\). Again, the derivative of \(\frac{3 \pi t}{2}\) is \(\frac{3 \pi}{2}\). Thus, the derivative of the cosine part is \(-\sin\left(\frac{3 \pi t}{2}\right) \cdot \frac{3 \pi}{2}\).
04

Sum the Derivatives

Combine the derivatives of each component: \(\frac{ds}{dt} = \frac{3 \pi}{2} \cos\left(\frac{3 \pi t}{2}\right) - \frac{3 \pi}{2} \sin\left(\frac{3 \pi t}{2}\right)\).
05

Simplify the Expression

Factor out the common term \(\frac{3 \pi}{2}\) from the expression to simplify it: \(\frac{ds}{dt} = \frac{3 \pi}{2} \left(\cos\left(\frac{3 \pi t}{2}\right) - \sin\left(\frac{3 \pi t}{2}\right)\right)\).

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Key Concepts

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

Understanding Trigonometric Functions
Trigonometric functions, such as sine and cosine, are at the heart of this exercise. These functions relate angles in a right triangle to the ratio of its sides. In the context of calculus, they are crucial for modeling periodic phenomena like waves and oscillations.
The sine function, denoted as \( \sin(x) \), oscillates between -1 and 1 as it traces the unit circle. Similarly, the cosine function, \( \cos(x) \), does the same but starts from 1 and follows the rotation. These functions are continuous and smooth, making them ideal candidates for differentiation.
When differentiating trigonometric functions, each function has a specific derivative:
  • The derivative of \( \sin(x) \) is \( \cos(x) \).
  • The derivative of \( \cos(x) \) is \(-\sin(x) \).
Understanding these derivatives allows us to tackle more complex functions that involve trigonometric components.
Applying the Chain Rule
The chain rule is a fundamental tool in calculus for differentiating composite functions. A composite function is a function made by combining two or more functions, like \( \sin(\frac{3 \pi t}{2}) \). Here, we apply the sine function to another function, \( \frac{3 \pi t}{2} \).

The chain rule provides a way to find the derivative of such a composite function. It states that the derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
Applying this to our exercise, the sine component involves:
  • \( f(x) = \sin(x) \) with \( f'(x) = \cos(x) \)
  • \( g(t) = \frac{3 \pi t}{2} \) with \( g'(t) = \frac{3 \pi}{2} \)
  • Thus, \( (f(g(t)))' = \cos(\frac{3 \pi t}{2}) \cdot \frac{3 \pi}{2} \)
By understanding and applying the chain rule, we effectively dissect complex functions into simpler parts, making differentiation manageable.
Introduction to Calculus and Differentiation
Calculus, the study of change, provides the tools needed to handle dynamic equations and model real-world changes. Differentiation, one of the core operations in calculus, allows us to determine the rate at which a function is changing at any point.
In simpler terms, if you think of a function as a road, differentiation tells us about the slope or steepness of the road at any given point. This is crucial in understanding how quickly or slowly a quantity changes over time.
In our example, we have a function combining sine and cosine—two periodic elements—affecting each other. Differentiation helps us understand how these interactions change over time. By simplifying derivatives into understandable units through rules like the chain rule, calculus demystifies the way we interpret and manage changes.
This foundational knowledge powers many scientific, engineering, and economic models, making calculus an indispensable part of these fields.

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Most popular questions from this chapter

Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time \(t\) sec is $$s=A \cos (2 \pi b t)$$' with \(A\) and \(b\) positive. The value of \(A\) is the amplitude of the motion, and \(b\) is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)

Area The area \(A\) of a triangle with sides of lengths \(a\) and \(b\) enclosing an angle of measure \(\theta\) is $$A=\frac{1}{2} a b \sin \theta.$$ a. How is \(d A / d t\) related to \(d \theta / d t\) if \(a\) and \(b\) are constant? b. How is \(d A / d t\) related to \(d \theta / d t\) and \(d a / d t\) if only \(b\) is constant? c. How is \(d A / d t\) related to \(d \theta / d t, d a / d t,\) and \(d b / d t\) if none of \(a, b,\) and \(\theta\) are constant?

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=\frac{4 x}{x^{2}+1}, \quad x_{0}=2$$

$$If r=\sin (f(t)), f(0)=\pi / 3, and f^{\prime}(0)=4, then what is d r / d t at t=0 ?$$

Ships Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

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