/*! 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 21 Find the slope of the curve at t... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the slope of the curve at the point indicated. $$y=\frac{1}{x-1}, \quad x=3$$

Short Answer

Expert verified
The slope is \(-\frac{1}{4}\).

Step by step solution

01

Identify the formula for finding the slope

The slope of a curve at a given point is found using the derivative of the function. The derivative represents the rate of change or the slope of the tangent line at any point on the curve. We need to find \( \frac{dy}{dx} \) for the given function.
02

Differentiate the function

To find the derivative of \( y=\frac{1}{x-1} \), we can use the power rule of differentiation and the chain rule. Rewriting \( y \) as \( (x-1)^{-1} \), the derivative \( \frac{dy}{dx} \) is given by:\[\frac{dy}{dx} = -1 \cdot (x-1)^{-2} \cdot 1 = -\frac{1}{(x-1)^2}\]
03

Substitute the x-value into the derivative

Now that we have \( \frac{dy}{dx} = -\frac{1}{(x-1)^2} \), substitute \( x = 3 \) into this expression to find the slope of the curve at that point:\[\frac{dy}{dx} = -\frac{1}{(3-1)^2} = -\frac{1}{4}\]
04

Interpret the result

The derivative \( \frac{dy}{dx} = -\frac{1}{4} \) represents the slope of the curve \( y = \frac{1}{x-1} \) at the point where \( x = 3 \). Hence, the slope of the tangent line to the curve at this point is \(-\frac{1}{4}\).

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Key Concepts

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

Derivative
When trying to understand what the derivative of a function is, it's helpful to think of it as a tool for understanding how a function behaves. The derivative tells us the rate at which a function is changing at any given point. - For example, if you think of a car traveling down a road, the derivative would represent the car's speed at any particular moment. - Mathematically, for a function y = f(x), the derivative is expressed as \( \frac{dy}{dx} \), meaning a small change in y (output) per unit change in x (input).By analyzing the derivative, we gain insights into many characteristics of the function, such as its increasing or decreasing nature and where it has peaks and valleys. This can be understood through the slope of the curve as well.
Power Rule
Differentiation often becomes much easier with rules like the power rule. The power rule is a basic yet powerful technique used in calculus to find derivatives quickly and efficiently.- In its simplest form, the power rule states: if \( y = x^n \) where n is any real number, then the derivative \( \frac{dy}{dx} = n \times x^{n-1} \).For example, if \( y = x^3 \), using the power rule we would differentiate to find \( \frac{dy}{dx} = 3x^2 \). This rule is quite handy when dealing with polynomial expressions. In our exercise, we initially expressed \( y = \frac{1}{x-1} \) as \( y = (x-1)^{-1} \) to make it easier to apply the power rule and chain rule.
Chain Rule
The chain rule in calculus is a method to differentiate composite functions, which are functions within functions. - Imagine you're evaluating a path in a movie theme park, where you choose one direction which leads you to another decision point—a bit like paths within paths.For a function y = f(g(x)), the derivative is given by the chain rule: \( \frac{dy}{dx} = f'(g(x)) \times g'(x) \). It's a way of peeling back the layers of functions to understand how they influence each other.In the exercise, the chain rule assists in differentiating \( y = (x-1)^{-1} \). Here, the outer function is \( f(u) = u^{-1} \) and the inner function is \( g(x) = x-1 \). This rule allowed us to compute the derivative \( \frac{dy}{dx} \) accurately.
Slope of Curve
The slope of a curve at a point provides a picture of its steepness and direction. When you hear about the slope, it typically refers to the change in y over the change in x, which is the essence of a derivative. - If the slope is positive, the curve ascends and increases from left to right; if negative, it descends. - A zero slope indicates a flat portion of the curve.In our given exercise, the slope at \( x = 3 \) was found by substituting into the derivative \( \frac{dy}{dx} = -\frac{1}{(x-1)^2} \), resulting in \( -\frac{1}{4} \). This negative slope shows the curve is decreasing at that specific point, giving a full picture of the tangent's orientation.

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Most popular questions from this chapter

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{3}-6 t^{2}+7 t, \quad 0 \leq t \leq 4$$

Verify that the following pairs of curves meet orthogonally. a. \(x^{2}+y^{2}=4, \quad x^{2}=3 y^{2}\) b. \(x=1-y^{2}, \quad x=\frac{1}{3} y^{2}\)

Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is \(1852 \mathrm{m}\) ). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u+\frac{1}{\cos ^{2} u}, \quad u=g(x)=\pi x, \quad x=1 / 4$$

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0)$$

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