91Ó°ÊÓ

Vector magnitude is a fundamental concept in integral calculus, especially when dealing with vectors. Imagine a vector as an arrow pointing in a specific direction with a certain length. This length is known as the vector's magnitude. To find the magnitude of a vector in two-dimensional space, we can use the formula:

• If a vector is written as \(\mathbf{r}(t) = x\mathbf{i} + y\mathbf{j}\) where \(x\) and \(y\) are functions of \(t\), then the magnitude \(\|\mathbf{r}(t)\|\) is given by \(\sqrt{x^2 + y^2}\).

Let's apply this to the given vector \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}\). Here, \(x = t\) and \(y = t^2\). Substituting into the formula gives:\[\|\mathbf{r}(t)\| = \sqrt{t^2 + (t^2)^2} = \sqrt{t^2 + t^4} = t\sqrt{1 + t^2}\]This expression represents the magnitude of the vector as a function of \(t\), giving us insight into how its length changes over the interval \([0, 3]\).

The substitution method is a powerful technique in integral calculus used to simplify complex integrals. This method relies on changing the variable of integration to make the integral easier to evaluate. Here's a step-by-step guide on using the substitution method:

• Identify a part of the integral that can be substituted. In our exercise, the term \(1 + t^2\) within \(t\sqrt{1 + t^2}\) is a natural choice.
• Set the substitution. Let \(u = 1 + t^2\). Consequently, differentiate to find \(du\): \(du = 2t\, dt\).
• Express \(t\, dt\) in terms of \(du\). Here, \(t\, dt = \frac{1}{2} du\).
• Change the limits of integration based on the substitution: When \(t = 0\), \(u = 1\). When \(t = 3\), \(u = 10\).

Plugging these into the integral transforms \(\int_{0}^{3} t \sqrt{1 + t^2} \, dt\) into \(\int_{1}^{10} \frac{1}{2} \sqrt{u} \, du\), significantly simplifying the integration process.

Calculating the curve length of a vector function is one of the many applications of integral calculus. This involves integrating the magnitude of the vector over a specified interval. For our exercise, finding the length of the curve is equivalent to evaluating the integral \(\int_{0}^{3} \|t \mathbf{i}+t^{2} \mathbf{j}\| \, dt\). Here's how you do it:

• First, compute the magnitude of the vector function over the interval. We found \(\|\mathbf{r}(t)\| = t \sqrt{1 + t^2}\).
• Set up the integral for the length of the curve. The length \(L\) can be found by integrally evaluating this magnitude over \([0, 3]\): \(L = \int_{0}^{3} t \sqrt{1 + t^2} \, dt\).
• Use the substitution method, as explained earlier, transforming and then solving the integral: \(\int_{1}^{10} \sqrt{u} \frac{1}{2} \, du\).
• Evaluate the transformed integral to find \(L = \frac{1}{3} (31.622 - 1) \) leading to an approximate value of 10.207.

This value, 10.207, represents the length of the vector curve from \(t = 0\) to \(t = 3\), showcasing how integral calculus helps in deriving such physical interpretations.