Chapter 18: Problem 5
(a) Find a linear approximation, \(p_{1}(t)\), to \(R(t)=\frac{1}{t}\) about \(t=0.5\). (b) Evaluate \(R(0.7)\) and \(p_{1}(0.7)\).
Short Answer
Expert verified
(a) The linear approximation is \(p_1(t) = 4 - 4t\). (b) \(R(0.7) \approx 1.4286\) and \(p_1(0.7) = 1.2\).
Step by step solution
01
Understand the Problem
We are tasked with finding a linear approximation, also known as the tangent line, to the function \(R(t) = \frac{1}{t}\) around the point \(t = 0.5\). We will then evaluate both the original function \(R(t)\) and the linear approximation \(p_1(t)\) at \(t = 0.7\).
02
Find the Derivative of R(t)
To create the linear approximation, we need the derivative of \(R(t)\). The function is given by \(R(t) = \frac{1}{t}\). The derivative \(R'(t)\) is found using the power rule: \(R'(t) = -\frac{1}{t^2}\).
03
Calculate R(t) and R'(t) at t=0.5
Substitute \(t = 0.5\) into both \(R(t)\) and \(R'(t)\):1. \(R(0.5) = \frac{1}{0.5} = 2\).2. \(R'(0.5) = -\frac{1}{(0.5)^2} = -4\).
04
Formulate the Linear Approximation p_1(t)
The formula for a linear approximation (tangent line) at \(t = a\) is given by:\[p_1(t) = R(a) + R'(a)(t - a)\]Substitute \(a = 0.5\), \(R(0.5) = 2\), and \(R'(0.5) = -4\):\[p_1(t) = 2 - 4(t - 0.5)\]Simplify to:\[p_1(t) = 2 - 4t + 2 = 4 - 4t\]
05
Evaluate R(0.7)
Substitute \(t = 0.7\) into the original function:\[R(0.7) = \frac{1}{0.7}\]Calculate using division:\[R(0.7) \approx 1.4286\]
06
Evaluate p_1(0.7)
Substitute \(t = 0.7\) into the linear approximation:\[p_1(0.7) = 4 - 4(0.7)\]Calculate:\[p_1(0.7) = 4 - 2.8 = 1.2\]
07
Compare R(0.7) and p_1(0.7)
The original function at \(t = 0.7\) gives \(R(0.7) \approx 1.4286\), while the linear approximation gives \(p_1(0.7) = 1.2\). This difference is due to the fact that linear approximations provide good estimates near the point of tangency and become less accurate further away.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Approximation
Linear approximation is a method used in calculus to estimate the value of a function around a specific point. It utilizes the concept of a tangent line to approximate the function. For a function \(f(t)\) near a point \(t = a\), the linear approximation is expressed as:
- \(p_1(t) = f(a) + f'(a)(t-a)\)
Derivatives
Derivatives play a crucial role in calculus, especially when it comes to finding linear approximations. The derivative of a function measures how the function's value changes as its input changes. It's essentially the slope of the tangent line at any given point. For the function \(R(t) = \frac{1}{t}\), the derivative is determined by using the power rule, resulting in:
- \(R'(t) = -\frac{1}{t^2}\)
Tangent Line
The tangent line is a straight line that just "touches" the function at a given point. It doesn't cross it at that location, and it's essentially the line that best represents the slope of the function at that precise point. For the function \(R(t) = \frac{1}{t}\), the tangent line at \(t = 0.5\) is formulated using the derivative and the value of \(R(t)\) at this point. The formula for the tangent line, also the linear approximation, is:
- \(p_1(t) = R(a) + R'(a)(t-a)\)
Function Evaluation
Function evaluation is the process of determining the value of a function at a specific input. In the exercise, we need to evaluate \(R(t) = \frac{1}{t}\) and its linear approximation \(p_1(t)\) at \(t = 0.7\). This involves substituting \(t = 0.7\) into the functions:
- \(R(0.7) = \frac{1}{0.7} \approx 1.4286\)
- \(p_1(0.7) = 4 - 4 \times 0.7 = 1.2\)