/*! 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 62 In Exercises \(51-70,\) find \(d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 62

In Exercises \(51-70,\) find \(d y / d t\). $$y=\cos \left(5 \sin \left(\frac{t}{3}\right)\right)$$

Short Answer

Expert verified
\( \frac{dy}{dt} = -\frac{5}{3} \cdot \sin(5 \sin(\frac{t}{3})) \cdot \cos(\frac{t}{3}) \)."}

Step by step solution

01

Understand the Function Composition

The given function is a composition of multiple functions. First, there's a sine function applied to \(\frac{t}{3}\), then that result is passed into another sine function, and finally, the cosine function is applied. Thus, we have three layers of functions: \(y = \cos(u)\), where \(u = 5 \cdot \sin(v)\), and \(v = \frac{t}{3}\).
02

Differentiate the Outer Function

Differentiate the outermost function using the chain rule. The derivative of \(y = \cos(u)\) with respect to \(u\) is \(-\sin(u)\). So, \( \frac{dy}{du} = -\sin(u)\).
03

Differentiate the Middle Function

Now differentiate the middle function \(u = 5 \cdot \sin(v)\) with respect to \(v\). The derivative is \( \frac{du}{dv} = 5 \cdot \cos(v)\).
04

Differentiate the Innermost Function

Differentiate the innermost function \(v = \frac{t}{3}\) with respect to \(t\). The derivative is \( \frac{dv}{dt} = \frac{1}{3}\).
05

Apply the Chain Rule

To find \( \frac{dy}{dt}\), we apply the chain rule to combine all these derivatives: \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt} = -\sin(u) \cdot 5 \cdot \cos(v) \cdot \frac{1}{3} \).
06

Substitute the Expressions Back

Substitute \(u = 5 \sin(v)\) and \(v = \frac{t}{3}\) back into the expression: \( \frac{dy}{dt} = -\sin(5 \sin(\frac{t}{3})) \cdot 5 \cdot \cos(\frac{t}{3}) \cdot \frac{1}{3}\).
07

Simplify the Expression

Simplify \( \frac{dy}{dt}\): \( \frac{dy}{dt} = -\frac{5}{3} \cdot \sin(5 \sin(\frac{t}{3})) \cdot \cos(\frac{t}{3}) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Composition of Functions
When dealing with the composition of functions, you're essentially working with a function inside another function. This can be thought of like layers of an onion. In the given exercise, we have three layers: the outer cosine function, an inner sine function, and the innermost being a simple linear expression \( t/3 \).
  • The innermost function, \( v = \frac{t}{3} \), is evaluated first, transforming \( t \).
  • The result, \( v \), is then plugged into the trigonometric function \( u = 5 \cdot \sin(v) \), creating another layer.
  • Finally, this result, \( u \), enters the final layer with the function \( y = \cos(u) \).
Understanding how these functions layer upon each other is crucial, especially when you apply differentiation rules like the chain rule.
Trigonometric Differentiation
Trigonometric differentiation deals with finding the derivatives of trigonometric functions, which are common in calculus problems. In this exercise, both the sine and cosine functions are involved.
**How to Differentiate Sine and Cosine:**
  • The derivative of \( \sin(x) \) with respect to \( x \) is \( \cos(x) \).
  • The derivative of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \).
For the given function \( y = \cos(5 \sin(\frac{t}{3})) \):
  • We first differentiate the cosine function, resulting in \( -\sin(u) \).
  • Then, the sine function inside needs differentiation, giving \( 5 \cdot \cos(v) \) where \( v = \frac{t}{3} \).
By carefully applying these rules, you can handle the trigonometric aspects of differentiation effectively.
Calculus Problem Solving
Solving calculus problems often involves recognizing patterns and functions within functions to properly apply differentiation rules. Let's break down this common calculus approach:
**Step-by-Step Application:**
  • Identify all the functions within the problem statement.
  • Differentiate from the outermost to the innermost function systematically using the chain rule.
In this example:
  • We start with the differentiation of the outer cosine layer, using the chain rule to manage each subsequent layer.
  • The process involves multiplying derivatives step by step: from the outermost, \( -\sin(u) \), to middle \( 5 \cdot \cos(v) \), and the inner \( \frac{1}{3} \).
  • Finally, substitute back to ensure your expression represents the rate of change correctly.
By taking these organized steps, calculus problems become much more approachable and manageable, allowing for more accurate problem solving.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\tan \theta) \sqrt{2 \theta+1}$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{2} 5 \theta$$

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 the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right)\) d. Find the 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. $$y=x^{3}-3 x^{2}-1, \quad 2 \leq x \leq 5, \quad x_{0}=\frac{27}{10}$$

A highway patrol plane flies 3 mi above a level, straight road at a steady \(120 \mathrm{mi} / \mathrm{h}\). The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi, the line-of-sight distance is decreasing at the rate of \(160 \mathrm{mi} / \mathrm{h}\). Find the car's speed along the highway.

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of \(\Delta T\), we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\)a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.