Problem 16
Question
Let \(f(x)=\frac{1}{x^{2}}\). Let \(P\) and \(Q\) be points on the graph of \(f\) with coordinates \((x, f(x))\) and \((x+\Delta x, f(x+\Delta x))\), respectively. (a) Find the slope of the secant line through \(P\) and \(Q\). Simplify your answer as much as possible. (b) By calculating the appropriate limit, nd the slope of the tangent line to \(f(x)\) at point \(P\)
Step-by-Step Solution
Verified Answer
The slope of the secant line through points P and Q on the graph of the function is \(-\frac{2x+ \Delta x}{{x}^{2}(x+ \Delta x)^{2}}\). The slope of the tangent line at point P is \(- \frac{2}{x^{3}}\).
1Step 1: Calculate the Difference Quotient
The slope of the secant line PQ is given by the difference quotient. This is defined as \(\frac{f(x+\Delta x) - f(x)}{\Delta x}\). Replace \(f(x)\) and \(f(x+\Delta x)\) with \( \frac{1}{x^{2}}\) and \( \frac{1}{(x+\Delta x)^{2}}\) respectively.
2Step 2: Simplify the Difference Quotient
In order to simplify the expression \(\frac{ \frac{1}{{x+\Delta x}^{2}} - \frac{1}{{x}^{2}}}{\Delta x}\), use the difference of squares formula and the common denominator to find \(- \frac{2x+ \Delta x}{{x}^{2}(x+ \Delta x)^{2}}\)
3Step 3: Find the Limit as \(\Delta x\) Approaches 0
Now, to find the slope of the tangent line to \(f(x)\) at point P, we need to calculate the limit of our difference quotient as \(\Delta x\) approaches 0 i.e., \(\lim_{\Delta x \to 0} - \frac{2x + \Delta x}{x^{2}(x + \Delta x)^{2}}\). After replacing \(\Delta x\) with 0 in this expression, we'll get \(- \frac{2}{x^{3}}\)
Key Concepts
DifferentiationLimitsRate of Change
Differentiation
Differentiation is a fundamental concept in calculus, allowing us to understand how functions change. It's all about finding the derivative of a function, which is essentially the rate at which the function's value changes as the input changes.
In simpler terms, it answers the question: "How do small changes in one variable affect changes in another variable?"
To differentiate a function like our given one, \(f(x) = \frac{1}{x^2}\), we use the concept of the derivative. This gives us the slope of the tangent line at any point on the function. The tangent line touches the graph at just one point, showing the instantaneous rate of change at that point.
In simpler terms, it answers the question: "How do small changes in one variable affect changes in another variable?"
To differentiate a function like our given one, \(f(x) = \frac{1}{x^2}\), we use the concept of the derivative. This gives us the slope of the tangent line at any point on the function. The tangent line touches the graph at just one point, showing the instantaneous rate of change at that point.
- The process starts with the difference quotient: \frac{f(x + \Delta x) - f(x)}{\Delta x}\.
- We simplify this quotient to reflect the function's actual rate of change.
- Finally, we apply limits to find an exact value, which leads to our derivative.
Limits
Limits help us discover what happens to a function as it approaches a particular point. In calculus, they're crucial for defining derivatives and integrals.
When solving for the slope of the tangent line in our exercise, we explored the limit as \(\Delta x o 0\). This tells us what's happening "right there" at that point on the curve.
When solving for the slope of the tangent line in our exercise, we explored the limit as \(\Delta x o 0\). This tells us what's happening "right there" at that point on the curve.
- A limit looks where the function is heading as the input changes close to a certain value.
- For example, in our problem, the expression \ \lim_{\Delta x \to 0} - \frac{2x + \Delta x}{x^2(x + \Delta x)^2}\ is simplified by setting \(\Delta x\) to 0, giving us precise behavior.
Rate of Change
The rate of change is a measure of how one quantity changes in relation to another. It's a key concept in understanding the dynamics of mathematical functions.
In the context of our exercise, the slope of the secant line gives us an average rate of change between two points \(P\) and \(Q\) on the curve. However, when we utilize limits to shrink the distance between these points, we find the slope of the tangent line. This gives us an instantaneous rate of change.
In the context of our exercise, the slope of the secant line gives us an average rate of change between two points \(P\) and \(Q\) on the curve. However, when we utilize limits to shrink the distance between these points, we find the slope of the tangent line. This gives us an instantaneous rate of change.
- The secant line represents a total change over an interval.
- The tangent line reveals how quickly values change at a single instant.
- For our function \(f(x) = \frac{1}{x^2}\), the tangent slope \(-\frac{2}{x^3}\) shows this exact rate at point \(P\).
Other exercises in this chapter
Problem 15
Let \(f(x)=2 x^{2}-5 x-1 .\) Let \(P\) and \(Q\) be points on the graph of \(f\) with \(x\) -coordinates 3 and \(3+h\), respectively. (a) Sketch the graph of \(
View solution Problem 16
Find the equation of the line tangent to \(y=\frac{2}{x+1}\) at the point \((1,1)\).
View solution Problem 17
We showed that the derivative of \(\sqrt{x}\) (or \(x^{\frac{1}{2}}\) ) is \(\frac{1}{2} \frac{1}{\sqrt{x}}\) (or \(\left.\frac{1}{2} x^{-\frac{1}{2}}\right) .\
View solution Problem 18
Let \(g\) be a function that is locally linear. We know that \(g^{\prime}(a)\) is the slope of the tangent line to the graph of \(g\) at point \(A=(a, g(a))\).
View solution