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Chapter 12: Problem 2

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

Short Answer

Expert verified
The integral evaluates to \(-3 \mathbf{i} + (4\sqrt{2} - 2) \mathbf{j} + 2 \mathbf{k}\).

Step by step solution

01

Understanding the Integral

The given integral is a vector integral where each component of the vector function is integrated separately with respect to the variable \( t \). Since it is in the form \( \int_{1}^{2} \left[ f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \right] dt \), we will evaluate \( \int_{1}^{2} f(t) dt \), \( \int_{1}^{2} g(t) dt \), and \( \int_{1}^{2} h(t) dt \) to find each component separately.
02

Compute the Integral for \( \mathbf{i} \)-Component

The \( \mathbf{i} \)-component of the integrand is \( 6 - 6t \). We need to find:\[\int_{1}^{2} (6 - 6t) \, dt\]Using the power rule for integration, it becomes: \[(6t - 3t^2) \Big|_{1}^{2}\]Evaluating at the limits gives us:\[[(6(2) - 3(2)^2) - (6(1) - 3(1)^2)] = (12 - 12) - (6 - 3) = 0 - 3 = -3\]
03

Compute the Integral for \( \mathbf{j} \)-Component

The \( \mathbf{j} \)-component is \( 3\sqrt{t} \). This requires us to evaluate:\[\int_{1}^{2} 3\sqrt{t} \, dt = \int_{1}^{2} 3t^{1/2} \, dt\]Using the power rule for integration, we have:\[3 \cdot \frac{t^{3/2}}{3/2} \Big|_{1}^{2} = 2t^{3/2} \Big|_{1}^{2}\]Evaluating this gives:\[[2(2)^{3/2} - 2(1)^{3/2}] = [2 \times 2\sqrt{2} - 2 \times 1] = 4\sqrt{2} - 2\]
04

Compute the Integral for \( \mathbf{k} \)-Component

For the \( \mathbf{k} \)-component, the function is \( \frac{4}{t^2} \). We compute:\[\int_{1}^{2} \frac{4}{t^2} \, dt = \int_{1}^{2} 4t^{-2} \, dt\]Again, using the power rule, we find:\[4 \cdot \frac{t^{-1}}{-1} \Big|_{1}^{2} = -4t^{-1} \Big|_{1}^{2}\]Evaluating this gives:\[[-4(1/2) + 4(1)] = [-2 + 4] = 2\]
05

Combine Results for the Final Vector

Now that we have computed each component, we combine them to get the result of the integral vector:\[(-3) \mathbf{i} + (4\sqrt{2} - 2) \mathbf{j} + (2) \mathbf{k}\]

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Key Concepts

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

Vector Integration
Vector integration involves integrating each component of a vector function separately. When you are given a function in the form \[\int \left[ f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \right] dt\]this means that you'll integrate the function for each vector component:
  • \[f(t)\] is integrated for the \(\mathbf{i}\) component.
  • \[g(t)\] is integrated for the \(\mathbf{j}\) component.
  • \[h(t)\] is integrated for the \(\mathbf{k}\) component.
Vector integration simplifies the process by allowing you to focus on individual integrals. For each part, integrate as you normally would in regular calculus, but keep track of the directions \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).When calculating the definite integral of a vector function from one boundary to another, remember:
  • Evaluate each component's antiderivative at the upper and lower limits.
  • Subtract the lower limit result from the upper limit for each component.
  • Combine these results, maintaining the vector direction indicators.
Power Rule for Integration
The power rule for integration is a handy tool for integrating functions of the form \(x^n\). Specifically, it states:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]for \(n eq -1\). Here, \(C\) is the constant of integration. This rule is straightforward to apply:
  • Increase the exponent by one.
  • Divide by the new exponent.
Using the power rule in definite integrals means you evaluate the antiderivative at specific limits. For \[\int_{a}^{b} x^n \, dx\], you compute:\[ \left( \frac{b^{n+1}}{n+1} \right) - \left( \frac{a^{n+1}}{n+1} \right) \]Applying this in vector problems:
  • Use the power rule on each vector component separately.
  • Evaluate and subtract, respecting the vector's directional components.
Remember, the power rule simplifies integration tasks in vector calculus by treating each part of the vector function independently.
Vector Function Components
When analyzing vector functions, it is essential to break them down into components aligned with \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\).Each component describes a part of the vector in a specific direction:
  • The \(\mathbf{i}\)-component describes the direction along the x-axis.
  • The \(\mathbf{j}\)-component describes the direction along the y-axis.
  • The \(\mathbf{k}\)-component describes the direction along the z-axis.
In a vector function \(\mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k}\), each function \(f(t), g(t),\) and \(h(t)\) takes a specific input \(t\) and determines the output magnitude along its respective axis.Understanding these components allows for detailed analysis of the vector's behavior in space. By integrating these components, you calculate how each part contributes to the overall change or motion of the vector over a given interval. This method forms the basis of solving complex vector calculus problems efficiently.

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Most popular questions from this chapter

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves in Exercises. $$\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0 < t < \pi / 2$$

Rounding the answers to four decimal places, use a CAS to find \(\mathbf{v}, \mathbf{a}\) speed, \(\mathbf{T}, \mathbf{N}, \mathbf{B}, \kappa,\) and the tangential and normal components of acceleration for the curves at the given values of \(t\) $$\mathbf{r}(t)=\left(3 t-t^{2}\right) \mathbf{i}+\left(3 t^{2}\right) \mathbf{j}+\left(3 t+t^{3}\right) \mathbf{k}, \quad t=1$$

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k}, \quad 0 \leq t \leq \pi$$

Use a CAS to perform the following steps. 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. $$\begin{aligned} &\mathbf{r}(t)=(\sin 2 t) \mathbf{i}+(\ln (1+t)) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 4 \pi,t_{0}=\pi / 4 \end{aligned}$$

Write a in the form \(\mathbf{a}=a_{\mathbf{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) without finding \(\mathbf{T}\) and \(\mathbf{N}\) $$\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+b t \mathbf{k}$$
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