/*! 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 4 Explain the relationships among ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Explain the relationships among the slope of a tangent line, the instantaneous rate of change, and the value of the derivative at a point.

Short Answer

Expert verified
Answer: The slope of a tangent line at a specific point on the graph of a function represents the instantaneous rate of change of the function at that point and is equal to the value of the derivative at that point. All three concepts are interconnected and describe the rate of change of the function at a specific point.

Step by step solution

01

1. Define the slope of a tangent line

The slope of a tangent line is the measure of the steepness of a line that touches a curve at exactly one point (tangent point). It measures the rate of change of the curve at the tangent point.
02

2. Define the instantaneous rate of change

The instantaneous rate of change is the rate of change of a function at a specific point. In other words, it measures how fast a variable is changing at that exact point. In terms of functions, it is the slope of the tangent line to the graph of the function at the given point.
03

3. Define the value of the derivative at a point

The derivative of a function represents the rate of change of the function with respect to its independent variable. The value of the derivative at a point is the instantaneous rate of change of the function at that specific point. This can be represented as f'(x) (read as f prime of x) if f(x) is the original function.
04

4. Relate the slope of a tangent line to the instantaneous rate of change

The slope of a tangent line at a specific point on the graph of a function and the instantaneous rate of change of that function at that point are the same. The slope of the tangent line measures the rate of change at the tangent point, which is also the value of the instantaneous rate of change.
05

5. Relate the instantaneous rate of change to the value of the derivative

The instantaneous rate of change of a function at a point is equivalent to the value of the derivative at that point. For the function f(x), if we want to find the instantaneous rate of change at a point x, we simply evaluate the derivative f'(x) at that specific point.
06

6. Establish the relationships among the three concepts

In summary, the slope of a tangent line, the instantaneous rate of change, and the value of the derivative at a point are all connected: - The slope of a tangent line at a specific point on the graph of a function represents the instantaneous rate of change of the function at that point. - The value of the derivative at a point is equal to the instantaneous rate of change of the function at that point. This implies that the slope of a tangent line is equal to the value of the derivative at that point on the function, which in turn measures the instantaneous rate of change.

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 91Ó°ÊÓ!

Key Concepts

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

Slope of a Tangent Line
When we look at a curve, we often want to understand how steep it is at a particular point. The slope of a tangent line helps us do just that. Picture a curve on a graph, and imagine a line that just skims the curve at a single point without cutting through it. The slope of this line reveals the steepness of the curve precisely at that point. It's like zooming in on the curve until it starts to look like a straight line, and then measuring the slope of that line.

Mathematically, the slope at any given point can be found using calculus. For a curve described by the function f(x), we find this slope by taking the derivative of the function, f'(x), and evaluating it at the point of interest. This slope is vital for understanding many physical phenomena, such as the acceleration of an object on a path at a specific moment.
Instantaneous Rate of Change
Now let's talk about how fast things are changing. The instantaneous rate of change is a way to describe how quickly a quantity, like position or temperature, changes at a particular instant. Think about driving a car and glancing at the speedometer; the speed shown is the instantaneous rate of how fast you're going at that second.

For functions, the instantaneous rate of change at a point is the derivative evaluated at that point, which is the same as the slope of the tangent line we just mentioned. This is a fundamental concept in motion and change, giving us a precise measurement of the rate at which something is changing right at that instant, without worrying about what happened before or what will happen next.
Value of the Derivative at a Point
If we know how to compute the instantaneous rate of change, we're already familiar with the value of the derivative at a point. Technically, the derivative of a function f(x) is another function, f'(x), that tells us the rate of change of f with respect to x at any point. When we say 'the value of the derivative at a point,' we mean the number that f'(x) gives us when we plug in a specific value for x. This number is crucial for solving problems in physics, engineering, and economics where the rate of change is tied to a specific instant.

It's important to recognize that while the derivative gives us a rate of change for every point, each 'value of the derivative at a point' is like a snapshot; it tells us the story of the function's behavior at that specific location along the x-axis.
Function Rate of Change
Exploring the broader picture, the function rate of change is how we talk about how a function behaves overall. This is not just at one point but across intervals or the entire domain of the function. This concept extends our understanding beyond the instant, showing us trends and patterns over time or space.

For a linear function, this rate of change is constant. But for more complex, nonlinear functions, the rate of change varies. The derivative of a function itself is a function that represents these rates of change at every possible point. By examining this derivative function, we can gain insight into the function's overall behavior, such as where it is increasing or decreasing, or where it has maxima or minima—critical information when analyzing the dynamics of systems in various scientific and mathematical applications.

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

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. The period \(T\) is the time required by the mass to complete one oscillation. Show that \(T=2 \pi \sqrt{\frac{m}{k}}\) b. Assume \(k\) is constant and calculate \(\frac{d T}{d m}\) c. Give a physical explanation of why \(\frac{d T}{d m}\) is positive.

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At \(40^{\circ}\) north latitude, the length of a day is approximated by $$D(t)=12-3 \cos \left(\frac{2 \pi(t+10)}{365}\right)$$ where \(D\) is measured in hours and \(0 \leq t \leq 365\) is measured in days, with \(t=0\) corresponding to January 1 a. Approximately how much daylight is there on March 1 \((t=59) ?\) b. Find the rate at which the daylight function changes. c. Find the rate at which the daylight function changes on March \(1 .\) Convert your answer to units of min/day and explain what this result means. d. Graph the function \(y=D^{\prime}(t)\) using a graphing utility. e. At what times of the year is the length of day changing most rapidly? Least rapidly?

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?

Tangent lines and exponentials Assume \(b\) is given with \(b > 0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, \(28,\) Mar 1997 )

See all solutions

Recommended explanations on Math Textbooks

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.