Problem 33 - Calculus III

Question

Evaluate \(\int_{0}^{3}\left\|t \mathbf{i}+t^{2} \mathbf{j}\right\| d t\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{1}{3} (10 \sqrt{10} - 1)\).

Step 1: Understanding the Problem

We need to evaluate the integral of a vector function's magnitude from 0 to 3. The function is given in terms of the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\).

Step 2: Find the Magnitude of the Vector

The vector given is \(t \mathbf{i} + t^2 \mathbf{j}\). The magnitude of this vector is \(\|t \mathbf{i} + t^2 \mathbf{j}\| = \sqrt{t^2 + (t^2)^2} = \sqrt{t^2 + t^4}\).

Step 3: Set Up the Integral

The integral we need to evaluate now becomes \(\int_{0}^{3} \sqrt{t^2 + t^4} \, dt\).

Step 4: Simplify the Integrand

Notice that \(t^2 + t^4 = t^2(1 + t^2)\). Therefore, \(\sqrt{t^2 + t^4} = \sqrt{t^2(1 + t^2)} = t\sqrt{1 + t^2}\).

Step 5: Evaluate the Integral

Now, we need to evaluate \(\int_{0}^{3} t\sqrt{1 + t^2} \, dt\). Use substitution: let \(u = 1 + t^2\), then \(du = 2t \, dt\) or \(t\, dt = \frac{1}{2} \, du\). Change the limits of integration from \(t\) to \(u\): when \(t=0, u=1\) and when \(t=3, u=10\). The integral becomes \(\frac{1}{2} \int_{1}^{10} \sqrt{u} \, du\).

Step 6: Perform the Integration

Integrate \(\sqrt{u}\): \(\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2}\). Then, evaluate from \(1\) to \(10\): \(\frac{1}{2} \left[ \frac{2}{3}(10)^{3/2} - \frac{2}{3}(1)^{3/2} \right]\).

Step 7: Calculate the Final Result

Calculate \((10)^{3/2}\) which is \(\sqrt{10^3} = \sqrt{1000} = 10 \sqrt{10}\), and \((1)^{3/2} = 1\). The result of the integral is \(\frac{1}{2} \left[ \frac{2}{3} \cdot 10 \sqrt{10} - \frac{2}{3} \cdot 1 \right] = \frac{1}{3} (10 \sqrt{10} - 1)\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that allow us to compute the accumulation of quantities over a particular interval. In essence, when we use definite integrals, we're finding the total change or area under a curve from one point to another. This is critical in vector calculus when dealing with vector functions over a specified range.
In our exercise, we are tasked with finding the definite integral of the magnitude of a vector from 0 to 3. This involves integrating across a defined interval, which can often give insights into the behavior of the system being analyzed.

• Definite integral is represented as \( \int_{a}^{b} f(x) \, dx \).
• The limits of integration \(a\) and \(b\) specify the interval.
• Provides a numerical result indicating total accumulation.

The integral \( \int_{0}^{3} \left\|t \mathbf{i}+t^{2} \mathbf{j}\right\| \, dt \) tells us how the magnitude of this vector behaves as \(t\) moves from 0 to 3.

Magnitude of Vectors

The magnitude of a vector is a measure of its length or size. In two-dimensional vector space, a vector has two components, corresponding to the standard unit vectors \( \mathbf{i} \) and \( \mathbf{j} \) in the Cartesian coordinate system. To find the magnitude of a vector, we simply apply the Pythagorean theorem to these components.
In the given exercise, the vector \( t \mathbf{i} + t^2 \mathbf{j} \) has components along the \(x\) and \(y\) axes. The magnitude is calculated as follows:

• Calculate each component squared: \( t^2 \) and \( (t^2)^2 \).
• Add the squared components: \( t^2 + t^4 \).
• Take the square root of the sum: \( \sqrt{t^2 + t^4} \).

Understanding the magnitude is crucial for evaluating where and how strongly the vector points in space over any given interval.

Substitution Method

The substitution method is a powerful technique in integration used to simplify the process of finding the integral of more complex functions. It involves changing the variable in the integral to make the integration simpler or more straightforward. This method is highly beneficial when dealing with composite functions or when the derivative of a substitution is present in the integrand.
To solve the integral \(\int_{0}^{3} t\sqrt{1 + t^2} \, dt\), we used substitution:

• Substitute \( u = 1 + t^2 \), which simplifies the expression inside the square root.
• Consequently, \( du = 2t \, dt \), provides a relation to replace \( t \, dt \).
• Adjust the limits of integration according to \( u \): when \( t=0, u=1\) and when \( t=3, u=10\).

This transformation reduces the complexity of integrating \( t\sqrt{1 + t^2} \), facilitating easier computation of the definite integral.

Integral Evaluation

After setting up the simpler integral through substitution, the next step is to perform the actual evaluation. Integral evaluation consists of finding the antiderivative of a function and then substituting the limits of integration.
In our problem, the integral \( \int_{1}^{10} \sqrt{u} \, du \) is straightforward once substitution is completed. Recall:

• The antiderivative of \( u^{1/2} \) is \( \frac{2}{3} u^{3/2} \).
• Substitute the limits into the antiderivative: \( \frac{2}{3}[(10)^{3/2} - (1)^{3/2}] \).
• Calculate \( (10)^{3/2} = 10 \sqrt{10} \); \( (1)^{3/2} = 1 \).

This provides a numerical solution of \( \frac{1}{3} (10 \sqrt{10} - 1) \), representing the total accumulation of vector magnitude over the given interval. Evaluating an integral like this informs us about geometric and physical properties of systems in vector calculus.