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Evaluate the integrals in Exercises \(21-26\) $$ \int_{1}^{4}\left[\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2 t} \mathbf{k}\right] d t $$

Short Answer

Expert verified
The integral evaluates to \( \ln 4 \mathbf{i} + \ln 4 \mathbf{j} + \frac{1}{2} \ln 4 \mathbf{k} \).

Step by step solution

01

Interpret the Integral

The given integral is a vector integral: \[ \int_{1}^{4}\left[\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2t} \mathbf{k}\right] dt \]This implies integrating each of the three components separately over the interval from 1 to 4.
02

Integrate the i-component

We find the integral of \( \frac{1}{t} \mathbf{i} \). The integral is:\[ \int \frac{1}{t} \ dt = \ln |t| + C \]Evaluate this from 1 to 4:\[ \left[ \ln |t| \right]_1^4 = \ln 4 - \ln 1 = \ln 4 \]
03

Integrate the j-component

Next, we integrate \( \frac{1}{5-t} \mathbf{j} \). The integral is:\[ \int \frac{1}{5-t} \ dt = -\ln |5-t| + C \]Evaluating from 1 to 4:\[ \left[ -\ln |5-t| \right]_1^4 = -\ln(5-4) + \ln(5-1) = -\ln 1 + \ln 4 = \ln 4 \]
04

Integrate the k-component

Now, integrate \( \frac{1}{2t} \mathbf{k} \). The integral is:\[ \int \frac{1}{2t} \ dt = \frac{1}{2} \ln |t| + C \]Evaluate from 1 to 4:\[ \left[ \frac{1}{2} \ln |t| \right]_1^4 = \frac{1}{2}(\ln 4 - \ln 1) = \frac{1}{2} \ln 4 \]
05

Combine Results

Combine the results from each component:- i-component: \( \ln 4 \mathbf{i} \)- j-component: \( \ln 4 \mathbf{j} \)- k-component: \( \frac{1}{2} \ln 4 \mathbf{k} \)Thus, the integral evaluates to:\[ \ln 4 \mathbf{i} + \ln 4 \mathbf{j} + \frac{1}{2} \ln 4 \mathbf{k} \]
06

Conclusion: Present the Integral Result

The solution to the integral is:\[ \ln 4 \mathbf{i} + \ln 4 \mathbf{j} + \frac{1}{2} \ln 4 \mathbf{k} \] which is the final result.

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Key Concepts

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

Vector Calculus
Vector calculus is a branch of mathematics that deals with vector fields and differentiable vector functions. It extends concepts from single-variable calculus, such as the integral and derivative, to multi-dimensional vectors.
When working with vectors, we often deal with quantities like displacement, velocity, and force, which have both magnitude and direction. In vector calculus, these are typically expressed in terms of unit vectors, such as \(\mathbf{i}, \mathbf{j}, \text{and } \mathbf{k} \), representing x, y, and z-components respectively.

  • **Vector Fields**: These are mathematical representations that assign a vector to every point in a given space, like how a magnetic field is visualized.
  • **Gradient, Divergence, and Curl**: Key operations in vector calculus that help describe how vectors change in a field. The gradient points in the direction of the greatest rate of increase of a scalar field.
In our problem, we address a vector valued function \(\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2t} \mathbf{k}\). Each component tells us about the behavior in its respective dimension. This understanding is crucial when transitioning from scalar calculus to vector calculus.
Definite Integrals
Definite integrals are used to calculate the net accumulation of a quantity and are fundamental to both scalar and vector calculus. A definite integral is represented as \(\int_{a}^{b} f(x) \, dx\) and computes the accumulation of \(f(x)\) from \(x=a\) to \(x=b\).

In vector calculus, definite integrals are often applied to vector functions, providing a cumulative sum of all instantaneous values of the vector function over an interval.

  • **Boundaries**: The limits \(a\) and \(b\) define the interval of integration.
  • **Evaluation**: After finding the antiderivative, the Fundamental Theorem of Calculus helps evaluate the expression at these limits.
In the given exercise, we evaluate the vector function over the interval \([1, 4]\). We treat each vector component as a separate integral. This gives us three definite integrals that we must solve to find the entire vector integral result.
Integral of Vector Functions
The integration of vector functions involves integrating each component separately. This is a straightforward application of definite integrals to each part of a vector.
Vector integrals often arise in physical contexts, such as calculating work done by a force or fluid flow along a path.

  • **Component-wise Integration**: Involves solving the integral of each component function independently, as in the given problem.
  • **Summing Results**: The final vector integral result is achieved by combining results from each component integration.
For the given integral:
  • \( \int_{1}^{4} \frac{1}{t} \, dt \) yields \( \ln 4 \mathbf{i} \)
  • \( \int_{1}^{4} \frac{1}{5-t} \, dt \) yields \( \ln 4 \mathbf{j} \)
  • \( \int_{1}^{4} \frac{1}{2t} \, dt \) yields \( \frac{1}{2} \ln 4 \mathbf{k} \)
The final integral expresses the cumulative change in each vector component from \(t=1\) to \(t=4\). The combined result: \( \ln 4 \mathbf{i} + \ln 4 \mathbf{j} + \frac{1}{2} \ln 4 \mathbf{k} \), encapsulates the entire vector path described by the original function.

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Most popular questions from this chapter

Ellipse a. Show that the curve \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(1-\cos t) \mathbf{k}\) \(0 \leq t \leq 2 \pi,\) is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane. b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at \(t=0, \pi / 2, \pi,\) and \(3 \pi / 2 .\) c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for \(t=0, \pi / 2, \pi,\) and 3\(\pi / 2\) to your sketch. d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary. e. Numerical integrator Estimate the length of the ellipse to two decimal places.

Use a CAS to perform the following steps in Exercises \(58-61 .\) a. Plot the space curve traced out by the position vector \(\mathbf{r}\) . b. Find the components of the velocity vector \(d \mathbf{r} / d t\) . c. Evaluate \(d \mathbf{r} / d t\) at the given point \(t_{0}\) and determine the equation of the tangent line to the curve at \(\mathbf{r}\left(t_{0}\right) .\) d. Plot the tangent line together with the curve over the given interval. $$ \begin{array}{l}{\mathbf{r}(t)=\left(\ln \left(t^{2}+2\right)\right) \mathbf{i}+\left(\tan ^{-1} 3 t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}} \\\ {-3 \leq t \leq 5, \quad t_{0}=3}\end{array} $$

Products of scalar and vector functions Suppose that the scalar function \(u(t)\) and the vector function \(\mathbf{r}(t)\) are both defined for \(a \leq t \leq b\) . a. Show that \(u \mathbf{r}\) is continuous on \([a, b]\) if \(u\) and \(\mathbf{r}\) are continuous on \([a, b] .\) If \(u\) and \(\mathbf{r}\) are both differentiable on \([a, b],\) show that \(u \mathbf{r}\) is differentiable on \([a, b]\) and that $$ \frac{d}{d t}(u \mathbf{r})=u \frac{d \mathbf{r}}{d t}+\mathbf{r} \frac{d u}{d t} $$

Evaluate the integrals in Exercises \(21-26\) $$ \int_{0}^{1}\left[t^{3} \mathbf{i}+7 \mathbf{j}+(t+1) \mathbf{k}\right] d t $$

Evaluate the integrals in Exercises \(21-26\) $$ \int_{1}^{2}\left[(6-6 t) \mathbf{i}+3 \sqrt{t} \mathbf{j}+\left(\frac{4}{t^{2}}\right) \mathbf{k}\right] d t $$

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