Problem 38
Question
(a) Let \(y=\sqrt{x} .\) Find \(d y\) and \(\Delta y\) at \(x=9\) with \(d x=\Delta x=-1\). (b) Sketch the graph of \(y=\sqrt{x},\) showing \(d y\) and \(\Delta y\) in the picture.
Step-by-Step Solution
Verified Answer
(a) At \(x=9\), \(dy=-\frac{1}{6},\Delta y=2\sqrt{2}-3\). (b) Graph shows tangent approximation.
1Step 1: Differentiate the Function
To find \(dy\), we need to differentiate \(y=\sqrt{x}\) with respect to \(x\). The derivative is \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\).
2Step 2: Evaluate the Derivative at Given Point
Substitute \(x = 9\) into the derivative to find \(\frac{dy}{dx}\) at this point: \[\frac{dy}{dx}\bigg|_{x=9} = \frac{1}{2\sqrt{9}} = \frac{1}{2 \times 3} = \frac{1}{6}.\]
3Step 3: Calculate dy
Use the differential formula \(dy = \frac{dy}{dx} \cdot dx\). Given \(dx = -1\), then \[dy = \frac{1}{6} \times (-1) = -\frac{1}{6}.\]
4Step 4: Calculate Change in y, Δy
To find \(\Delta y\), compute \(y\) at both \(x = 9\) and \(x = 8\). First, \(y = \sqrt{9} = 3\). Next, \(x = 8\), so \(y = \sqrt{8} = 2\sqrt{2}\). Calculate \(\Delta y = y(x=8) - y(x=9) = 2\sqrt{2} - 3\).
5Step 5: Sketch the Graph and Show dy and Δy
Draw the graph of \(y = \sqrt{x}\), marking the points for \(x = 9\) and \(x = 8\). The tangent line at \(x = 9\) shows \(dy\) as a small linear increment, and the vertical drop from point \(x=9\) to \(x=8\) on the curve represents \(\Delta y\). This highlights the difference between the linear approximation and the actual change on the graph.
Key Concepts
derivativedifferentialsgraph sketching
derivative
A derivative represents the rate of change of a function concerning its variable, essentially providing a "slope" at any point. In this problem, we start by differentiating the function given by \(y=\sqrt{x}\). The result is the derivative \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\). This formula means for every small change in \(x\), there is a corresponding change in \(y\) that follows this ratio as \(x\) varies.
This derivative is vital because it predicts how \(y\) will change with \(x\). In our scenario, evaluating it at \(x=9\) gives us \(\frac{1}{6}\). This value tells us precisely the steepness of the curve at \(x=9\), helping us know whether \(y\) is increasing or decreasing at that point.
Understanding the derivative as a concept allows you to visualize changes along the curve. When calculating derivatives, remember they provide instant rates of change, giving powerful insight into the behavior of curves everywhere.
This derivative is vital because it predicts how \(y\) will change with \(x\). In our scenario, evaluating it at \(x=9\) gives us \(\frac{1}{6}\). This value tells us precisely the steepness of the curve at \(x=9\), helping us know whether \(y\) is increasing or decreasing at that point.
Understanding the derivative as a concept allows you to visualize changes along the curve. When calculating derivatives, remember they provide instant rates of change, giving powerful insight into the behavior of curves everywhere.
- First, identify the function you need to differentiate.
- Apply rules of differentiation to find \(\frac{dy}{dx}\).
- Evaluate \(\frac{dy}{dx}\) at the specific value of \(x\).
differentials
Differentials, like \(dy\), are small differences that offer a linear approximation of changes in \(y\), given a change in \(x\), \(dx\). These differ from \(\Delta y\), which denotes the actual change in \(y\) when \(x\) changes from one value to another. Think of differentials as providing a tangent line's slope over a small interval, while \(\Delta y\) measures the exact vertical distance between two points on the curve.
For the function \(y=\sqrt{x}\), at \(x=9\) when \(dx=-1\), we calculated \(dy\) by using the formula \(dy = \frac{dy}{dx} \cdot dx = -\frac{1}{6}\). This means as \(x\) decreases by 1 unit, \(y\) decreases approximately by \(\frac{1}{6}\) under linear approximation.
In contrast, \(\Delta y\) is computed from the value of \(y\) at \(x=9\) and \(x=8\). Performing this calculation showed a change of \(2\sqrt{2} - 3\). This highlights the difference between these two concepts: \(dy\) provides a linear approximation, perfect to understand the immediate change, while \(\Delta y\) describes the total change across the interval.
For the function \(y=\sqrt{x}\), at \(x=9\) when \(dx=-1\), we calculated \(dy\) by using the formula \(dy = \frac{dy}{dx} \cdot dx = -\frac{1}{6}\). This means as \(x\) decreases by 1 unit, \(y\) decreases approximately by \(\frac{1}{6}\) under linear approximation.
In contrast, \(\Delta y\) is computed from the value of \(y\) at \(x=9\) and \(x=8\). Performing this calculation showed a change of \(2\sqrt{2} - 3\). This highlights the difference between these two concepts: \(dy\) provides a linear approximation, perfect to understand the immediate change, while \(\Delta y\) describes the total change across the interval.
- Use \(dy = \frac{dy}{dx} \cdot dx\) to estimate small changes linearly.
- Find \(\Delta y = y(x+n) - y(x)\) for the actual change over an interval.
graph sketching
Graph sketching visually communicates the behavior of functions, making calculus concepts like \(dy\) and \(\Delta y\) easier to grasp. To illustrate changes, we sketch the curve \(y=\sqrt{x}\). At \(x=9\), plot the function's graph, and draw a tangent representing the derivative \(\frac{1}{6}\), showing the instant slope or rate of change.
This tangent line allows us to visualize \(dy\) — our linear approximation. On the graph, this looks like a small step down along the tangent. In contrast, \(\Delta y\) shows as the actual drop from the curve at \(x=9\) to \(x=8\). These graphical elements help conceptualize the difference between the approximated change \(dy\) and the precise change \(\Delta y\).
Graph sketeching is a powerful tool in differential calculus, making it easier to interpret analytical results.
This tangent line allows us to visualize \(dy\) — our linear approximation. On the graph, this looks like a small step down along the tangent. In contrast, \(\Delta y\) shows as the actual drop from the curve at \(x=9\) to \(x=8\). These graphical elements help conceptualize the difference between the approximated change \(dy\) and the precise change \(\Delta y\).
Graph sketeching is a powerful tool in differential calculus, making it easier to interpret analytical results.
- Start by plotting the function on a coordinate plane.
- Add the tangent line at the point for visualizing \(dy\).
- Illustrate the vertical difference for \(\Delta y\).
Other exercises in this chapter
Problem 38
Find the limits. $$\lim _{x \rightarrow+\infty}[\ln x-\ln (1+x)]$$
View solution Problem 38
A point \(P\) is moving along the curve whose equation is \(y=\sqrt{x^{3}+17} .\) When \(P\) is at \((2,5), y\) is increasing at the rate of 2 units/s. How fast
View solution Problem 38
At what point(s) is the tangent line to the curve \(y^{3}=2 x^{2}\) perpendicular to the line \(x+2 y-2=0 ?\)
View solution Problem 38
Find \(d y / d x\) using any method. $$y=\left(2 x^{2}-2 x+1\right) e^{2 x}$$
View solution