Question

Find the total mass of the wire with density \(\boldsymbol{\rho}\). \(\mathbf{r}(t)=t^{2} \mathbf{i}+2 t \mathbf{j}, \quad \rho(x, y)=\frac{3}{4} y, \quad 0 \leq t \leq 1\)

Short answer

The total mass of the wire is \( m = \frac{3}{2}\left(2 - \sqrt{2}\right) \)

Step-by-step solution

Recall the mass of an object with density

If the object under consideration has variable density, then its total mass \( m \) can be found by integrating the density \( \boldsymbol{\rho} \) over the entire volume \( V \) of the object:\[ m = \int_{V} \boldsymbol{\rho} \, dV \]where \( dV \) denotes a small volume element.

Calculate the line element

To express the volume element in terms of the 1D parameter \( t \), differentiate the given parametric equation\[ \mathbf{r}(t) = t^{2} \mathbf{i} + 2t \mathbf{j} \]with respect to \( t \) to get the tangent vector\[ \mathbf{r'}(t) = 2t \mathbf{i} + 2 \mathbf{j} \]Then, the magnitude of the tangent vector, denoted \( \left|\mathbf{r'}(t)\right| \), gives the length of a small line element along the curve. Thus\[ \left|\mathbf{r'}(t)\right| = \sqrt{(2t)^{2} + 2^{2}} = 2\sqrt{t^{2} + 1} \]

Calculate the total mass

Substitute the given density function\[ \boldsymbol{\rho}(x, y) = \frac{3}{4}y = \frac{3}{4}(2t) = \frac{3}{2}t \]and the value from step 2 into the mass formula from step 1 and integrate from \( t = 0 \) to \( t = 1 \):\[ m = \int_0^1 \frac{3}{2}t \cdot 2\sqrt{t^{2} + 1} \, dt = 3 \int_0^1 t\sqrt{t^{2} + 1} \, dt \]Completing the square in the integral gives\[ m = 3 \int_0^1 \frac{1}{2}d\left(t^{2} + 1\right) = 3\left[\frac{1}{2}\left(t^{2} + 1\right)^{\frac{3}{2}}\right]_0^1 = \frac{3}{2}\left(2 - \sqrt{2}\right) \]