Question

Sketch the plane curve and find its length over the given interval. $$ \mathbf{r}(t)=t \mathbf{i}+3 t \mathbf{j}, \quad[0,4] $$

Short answer

The sketch of the curve \mathbf{r}(t) is a straight line with slope 3 passing through the origin, and the length of the curve from t = 0 to t = 4 is 4\(\sqrt{10}\).

Step-by-step solution

Identify the component functions

The vector function representing the curve is \mathbf{r}(t) = \( t \mathbf{i} + 3t \mathbf{j}\). From this, we can see that the component functions are \(x(t) = t\) and \(y(t) = 3t\). These functions define the curve in the xy-plane.

Sketch the Curve

Sketch a curve on the xy-plane using the component functions from step 1. This curve is a straight line with slope 3 passing through the origin (as for every unit increase in x, y increases by 3 units).

Derive the component functions

Derive the component functions to get the rate of change of each function with respect to t. The derivative \(dx/dt = 1\) and the derivative \(dy/dt = 3\).

Compute the length of the curve

Use the formula for the length of a curve: \[L = \int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2} dt\]. Substituting our values, we get \[L = \int_0^4 \sqrt{1^2 + 3^2} dt\] = \(\int_0^4 \sqrt{10} dt\) = 4\(\sqrt{10}\).