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Chapter 3: Problem 72

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

Short Answer

Expert verified
\( \frac{dr}{dt} \) at \( t = 0 \) is 2.

Step by step solution

01

Identify the function and determine the need for the chain rule

We are given the function \( r = \sin(f(t)) \). Since \( f(t) \) is a function of \( t \), and \( r \) is a function of \( f(t) \), we need to use the chain rule to find the derivative \( \frac{dr}{dt} \).
02

Apply the chain rule

The chain rule states that \( \frac{dr}{dt} = \frac{dr}{df} \cdot \frac{df}{dt} \). First, find \( \frac{dr}{df} \); since \( r = \sin(f(t)) \), we have \( \frac{dr}{df} = \cos(f(t)) \).
03

Determine \( \frac{df}{dt} \) at \( t = 0 \)

Given that \( f'(0) = 4 \), we have \( \frac{df}{dt} \bigg|_{t=0} = 4 \).
04

Evaluate \( \cos(f(t)) \) at \( t = 0 \)

Since \( f(0) = \frac{\pi}{3} \), substitute this to get \( \cos(f(0)) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
05

Calculate \( \frac{dr}{dt} \) at \( t = 0 \)

Now substitute the values from Steps 2, 3, and 4 into the chain rule: \( \frac{dr}{dt} = \cos(f(0)) \cdot f'(0) = \frac{1}{2} \cdot 4 = 2 \).

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Key Concepts

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

Derivative Calculation
Calculating derivatives is a fundamental skill in calculus. It involves understanding how a function changes as its input changes.
  • In the exercise, we were asked to find the derivative of a variable, \( r \), with respect to time \( t \). This is denoted mathematically by \( \frac{dr}{dt} \).
  • Derivatives allow us to measure the rate of change of a quantity. For example, in this exercise, \( \frac{dr}{dt} \) gives us the rate at which \( r \) changes as \( t \) changes.
  • To calculate \( \frac{dr}{dt} \), we need to use the chain rule because the function involves a composition of functions.
This process involves identifying the relationships between different functions and using rules of differentiation to find the desired rate of change.
Trigonometric Functions
In calculus, trigonometric functions like sine and cosine often appear in problems involving rates of change.
  • The sine function in our problem, \( r = \sin(f(t)) \), represents one of the basic trigonometric relationships. The derivative of sine plays a crucial role in the solution.
  • Knowing that \( \frac{d}{df}(\sin(f)) = \cos(f) \) is important because it tells us how sine changes with its input. This idea helps break down the function into manageable parts for calculation.
  • Understanding trigonometric identities, such as \( \cos(\frac{\pi}{3}) = \frac{1}{2} \), allows us to evaluate trigonometric functions at specific points. This is a crucial step in applying the chain rule.
Trigonometric functions add complexity to calculus problems but are manageable when approached methodically.
Chain Rule Application
The chain rule is a vital tool in calculus for differentiating compositions of functions.
  • The chain rule formula is \( \frac{dr}{dt} = \frac{dr}{df} \cdot \frac{df}{dt} \). Here, it helps find the derivative of a function composed of other functions, such as \( \sin(f(t)) \).
  • First, you differentiate the outer function. In this problem, that's \( \sin(f(t)) \) with respect to \( f(t) \), giving \( \frac{dr}{df} = \cos(f(t)) \).
  • Next, differentiate the inner function, \( f(t) \), with respect to \( t \), which is provided as \( f'(0) = 4 \).
  • Finally, multiply these derivatives, substituting known values. The calculated \( \cos(f(0)) = \frac{1}{2} \) when multiplied by \( 4 \) gives the rate of change, \( 2 \).
The chain rule simplifies complex derivatives by breaking them into easier, smaller steps.

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Most popular questions from this chapter

Find \(y^{\prime \prime}\) $$y=(1-\sqrt{x})^{-1}$$

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 \(k T / 2\)

Find \(d y / d t\) $$y=\left(\frac{t^{2}}{t^{3}-4 t}\right)^{3}$$

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, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1$$

Assume that \(f^{\prime}(3)=-1, g^{\prime}(2)=5, g(2)=3,\) and \(y=f(g(x))\) What is \(y^{\prime}\) at \(x=2 ?\)
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