Chapter 2: Problem 2
Let \(y=2 x^{3}-x\). a. Find \(\Delta x\) and \(\Delta y\) if \(x\) changes from 2 to \(1.97\). b. Find the differential \(d y\), and use it to approximate \(\Delta y\) if \(x\) changes from 2 to \(1.97\). c. Compute \(\Delta y-d y\), the error in approximating \(\Delta y\) by \(d y\).
Short Answer
Expert verified
a. \(\Delta x = -0.03\) and \(\Delta y = -3.24483\)
b. \(dy = -0.69\)
c. \(\Delta y - dy = -2.55483\)
Step by step solution
01
The given function is \(y = 2x^3 - x\). To find its derivative with respect to \(x\), we will use the power rule: \[\frac{d}{dx}(x^n) = nx^{n-1}\] The derivative dy/dx (also written as \(y'\)) is: \[\frac{dy}{dx} = 6x^2 - 1\] Now we will compute the values for each part of the problem. #Step 2: Compute Δx and Δy#
We are given that \(x\) changes from 2 to 1.97:
\(\Delta x = x_1 - x_0 = 1.97 - 2 = -0.03\)
Now let's calculate \(\Delta y\):
\[\Delta y = y(x_1) - y(x_0)\]
First, compute the value of the function at \(x_1 = 1.97\) and \(x_0 = 2\):
\[y(1.97) = 2(1.97)^3 - 1.97 = 12.72517 - 1.97 = 10.75517\]
\[y(2) = 2(2)^3 - 2 = 16 - 2 = 14\]
Now, compute \(\Delta y\):
\[\Delta y = 10.75517 - 14 = -3.24483\]
#Step 3: Find the differential dy and approximate Δy#
02
Now we will find the differential \(dy\). Recall that the derivative we found in Step 1 is: \[\frac{dy}{dx} = 6x^2 - 1\] At \(x_0=2\), the derivative is: \[\frac{dy}{dx}(2) = 6(2)^2 - 1 = 24 - 1 = 23\] Now using the differential to approximate \(\Delta y\): \[dy = (\frac{dy}{dx}) \cdot \Delta x = 23 \cdot (-0.03) = -0.69\] #Step 4: Compute the error Δy - dy#
Now we will compute the error:
\[\Delta y - dy = -3.24483 - (-0.69) = -3.24483 + 0.69 = -2.55483\]
The error in approximating \(\Delta y\) by \(dy\) is -2.55483.
In summary:
a. \(\Delta x = -0.03\) and \(\Delta y = -3.24483\)
b. The differential \(dy = -0.69\) was used to approximate \(\Delta y\)
c. The error in approximating \(\Delta y\) by \(dy\) is -2.55483.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Delta x and Delta y
Understanding the change in variables is essential to grasp the fundamental concept of calculus. When a function's input value shifts from one point to another, the alteration in the variable is denoted as \(\text{Δ}x\). Similarly, the corresponding change in the function's output is represented by \(\text{Δ}y\). This provides us with a straightforward calculation of how much the function's output varies due to a change in the input.
For instance, if we consider the function \(\text{y} = 2x^3 - x\) and examine the input value shifting from 2 to 1.97, we calculate \(\text{Δ}x = 1.97 - 2 = -0.03\). To find \(\text{Δ}y\), we determine the output at both input points and then compute the difference: \(\text{Δ}y = y(1.97) - y(2)\). This yields the change in the function's output as -3.24483.
Visualizing this on a graph, \(\text{Δ}x\) and \(\text{Δ}y\) together form a straight line segment between the points on the curve of the function at \(\text{x = 2}\) and \(\text{x = 1.97}\). This line segment approximates the actual curve of the function over the small interval we are considering.
For instance, if we consider the function \(\text{y} = 2x^3 - x\) and examine the input value shifting from 2 to 1.97, we calculate \(\text{Δ}x = 1.97 - 2 = -0.03\). To find \(\text{Δ}y\), we determine the output at both input points and then compute the difference: \(\text{Δ}y = y(1.97) - y(2)\). This yields the change in the function's output as -3.24483.
Visualizing this on a graph, \(\text{Δ}x\) and \(\text{Δ}y\) together form a straight line segment between the points on the curve of the function at \(\text{x = 2}\) and \(\text{x = 1.97}\). This line segment approximates the actual curve of the function over the small interval we are considering.
Differentials in Calculus
Differentials in calculus, denoted as \(\text{dy}\) and \(\text{dx}\), serve as infinitesimally small changes in the function's output and input, respectively. The differential \(\text{dy}\), in particular, is the product of the derivative of the function and the differential \(\text{dx}\).
The derivative—found using rules like the power rule—gives us the slope of the tangent line to the curve of a function at any given point. When we multiply this slope by the small change in \(\text{x}\), represented as \(\text{Δ}x\), we obtain the differential \(\text{dy}\) which provides an excellent approximation of \(\text{Δ}y\) for small values of \(\text{Δ}x\). The formula for the differential is \(\text{dy} = (\frac{dy}{dx}) \times \text{Δ}x\).
In our case, for the function \(\text{y} = 2x^3 - x\), at \(\text{x}_0=2\), the calculated derivative is 23. Hence, the differential \(\text{dy} = 23 \times (-0.03) = -0.69\) suggests a linear approximation of the change in \(\text{y}\).
Using differentials, we can make predictions about a function's behavior, simplifying complex changes into manageable linear estimations, which are particularly powerful in engineering, physics, and economics for modeling small-scale variations.
The derivative—found using rules like the power rule—gives us the slope of the tangent line to the curve of a function at any given point. When we multiply this slope by the small change in \(\text{x}\), represented as \(\text{Δ}x\), we obtain the differential \(\text{dy}\) which provides an excellent approximation of \(\text{Δ}y\) for small values of \(\text{Δ}x\). The formula for the differential is \(\text{dy} = (\frac{dy}{dx}) \times \text{Δ}x\).
In our case, for the function \(\text{y} = 2x^3 - x\), at \(\text{x}_0=2\), the calculated derivative is 23. Hence, the differential \(\text{dy} = 23 \times (-0.03) = -0.69\) suggests a linear approximation of the change in \(\text{y}\).
Using differentials, we can make predictions about a function's behavior, simplifying complex changes into manageable linear estimations, which are particularly powerful in engineering, physics, and economics for modeling small-scale variations.
Linear Approximation Error
Linear approximation is a valuable method for estimating the value of a function near a given point using the function's derivative. However, it is not always perfectly accurate, and the discrepancy between the actual change in the function's output \(\text{Δ}y\) and the linear approximation provided by the differential \(\text{dy}\) is known as the linear approximation error.
The error can be quantified as \(\text{Δ}y - \text{dy}\). It is crucial for students to understand that this error typically increases with a larger interval between the two points or when the function's curvature is more pronounced within that interval. For small intervalls, as in our example from \(\text{x = 2}\) to \(\text{x = 1.97}\), the error in using \(\text{dy}\) to approximate \(\text{Δ}y\) is quite minimal, yielding a value of -2.55483 in this particular instance.
The error can be quantified as \(\text{Δ}y - \text{dy}\). It is crucial for students to understand that this error typically increases with a larger interval between the two points or when the function's curvature is more pronounced within that interval. For small intervalls, as in our example from \(\text{x = 2}\) to \(\text{x = 1.97}\), the error in using \(\text{dy}\) to approximate \(\text{Δ}y\) is quite minimal, yielding a value of -2.55483 in this particular instance.