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Magazine Chapter 3: Problem 71
Evaluate \(\int_{0}^{3}\left\|t \mathbf{i}+t^{2} \mathbf{j}\right\| d t\).
Short Answer
Expert verified
The integral evaluates to approximately 10.207.
Step by step solution
01
Understanding the vector function
The vector function given is \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j}\). The magnitude, or norm, of this vector is \(\|\mathbf{r}(t)\|\).
02
Calculate the magnitude of the vector function
The magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) is given by \(\sqrt{a^2 + b^2}\). Thus, the magnitude of \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j}\) is \(\sqrt{t^2 + (t^2)^2} = \sqrt{t^2 + t^4} = \sqrt{t^2(1 + t^2)}\).
03
Simplify the expression under the integral
The expression simplifies to \(|t|\sqrt{1 + t^2}\). Since \(t\) is between 0 and 3, \(|t| = t\), so the integral becomes \(\int_0^3 t\sqrt{1 + t^2} \, dt\).
04
Choose an appropriate substitution
Let's use the substitution \(u = 1 + t^2\), then \(du = 2t \, dt\), or \(t \, dt = \frac{1}{2} \, du\). We need to change the limits: when \(t = 0\), \(u = 1\), and when \(t = 3\), \(u = 10\). Substituting these expressions gives \(\int_1^{10} \frac{1}{2}\sqrt{u} \, du\).
05
Evaluate the integral
Now, evaluate \(\int_1^{10} \frac{1}{2} u^{1/2} \, du\). Integrate to get \(\frac{1}{2}\left[ \frac{2}{3} u^{3/2} \right]_1^{10}\). This simplifies to \(\left[ \frac{1}{3} u^{3/2} \right]_1^{10}\).
06
Compute the definite integral
Using the antiderivative, evaluate it from 1 to 10: \(\frac{1}{3}(10^{3/2}) - \frac{1}{3}(1^{3/2})\). Calculate this as \(\frac{1}{3}(10\sqrt{10}) - \frac{1}{3}(1) = \frac{1}{3}(31.622 - 1) = \frac{1}{3}(30.622) = 10.207\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Calculus
In the context of mathematics, vector calculus is a branch that deals with vector fields and lines of vector operations. A vector function such as \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j}\) is often analyzed through its direction and magnitude.
A vector is simply a quantity that has both magnitude (size) and direction.
A vector is simply a quantity that has both magnitude (size) and direction.
- Vectors can be represented in a plane using unit vectors \(\mathbf{i}\) and \(\mathbf{j}\),.
- In this case, the vector \(\mathbf{r}(t)\) changes as time \(t\) changes.
Integration by Substitution
Integration by substitution is a technique used to simplify integrals, making them more manageable.
In our exercise, we made use of substitution by letting \( u = 1 + t^2 \) and \( du = 2t \, dt \). This transforms a complicated integral into a simpler form.
In our exercise, we made use of substitution by letting \( u = 1 + t^2 \) and \( du = 2t \, dt \). This transforms a complicated integral into a simpler form.
- The idea is to replace \( t \) and \( dt \) with terms in \( u \) and \( du \).
- Changing the variable also involves changing the limits of the integral.
Magnitude of a Vector
The magnitude of a vector provides the "length" or "size" of the vector. To find the magnitude of a given vector function like \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \), we use the formula:
\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2} \]
In our case, we substitute \( a = t \) and \( b = t^2 \) to get:
\[ \|\mathbf{r}(t)\| = \sqrt{t^2 + t^4} = \sqrt{t^2(1 + t^2)} \]
\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2} \]
In our case, we substitute \( a = t \) and \( b = t^2 \) to get:
\[ \|\mathbf{r}(t)\| = \sqrt{t^2 + t^4} = \sqrt{t^2(1 + t^2)} \]
- Magnitude calculations are crucial when dealing with quantities like velocity or force in physics.
- They help in understanding the intensity or extent of the vector.
Definite Integrals
Definite integrals are used to calculate the area under a curve between two limits. Here, we evaluated the definite integral \( \int_0^3 t\sqrt{1 + t^2} \, dt \) which arose from the magnitude of the vector function.
The process involves calculating the antiderivative first and then finding the difference between its values at the upper and lower limits.
The process involves calculating the antiderivative first and then finding the difference between its values at the upper and lower limits.
- The result of a definite integral is a number that represents the net "accumulated value" (such as area, distance, or mass).
- It's important to change and apply the limits appropriately through substitution.
