Find the mass of a wire that lies along the curve r​(t)equals=left parenthesis t squared minus 6 right parenthesist2−6jplus+2tk​, 0 less than or equals t less than or equals 20≤t≤2​, if the density is deltaδequals=three halves t
3
2t.

To find the mass of a wire that lies along the curve \( \mathbf{r}(t) = (t^2 - 6)\mathbf{j} + 2t\mathbf{k} \), where \( 0 \leq t \leq 2 \), and the density is \( \delta(t) = \frac{3}{2}t \), we follow these steps:

### Step 1: Express the Curve in Vector Form
The curve is given by:
\[
\mathbf{r}(t) = (t^2 - 6)\mathbf{j} + 2t\mathbf{k}
\]
Here, the position vector \( \mathbf{r}(t) \) has no \( \mathbf{i} \)-component, only \( \mathbf{j} \) and \( \mathbf{k} \) components.

### Step 2: Compute the Differential Length \( ds \)
The differential arc length \( ds \) along the curve is given by:
\[
ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} dt
\]
Given \( \mathbf{r}(t) \), we find:
- \( x(t) = 0 \) (so \( \frac{dx}{dt} = 0 \))
- \( y(t) = t^2 - 6 \), hence \( \frac{dy}{dt} = 2t \)
- \( z(t) = 2t \), hence \( \frac{dz}{dt} = 2 \)

So, the differential length \( ds \) is:
\[
ds = \sqrt{(0)^2 + (2t)^2 + (2)^2} dt = \sqrt{4t^2 + 4} dt = \sqrt{4(t^2 + 1)} dt = 2\sqrt{t^2 + 1} dt
\]

### Step 3: Set Up the Mass Integral
The mass \( M \) of the wire is given by:
\[
M = \int_C \delta(t) \, ds
\]
Substitute the given density function \( \delta(t) = \frac{3}{2}t \) and the expression for \( ds \):
\[
M = \int_{0}^{2} \frac{3}{2}t \times 2\sqrt{t^2 + 1} \, dt = 3 \int_{0}^{2} t\sqrt{t^2 + 1} \, dt
\]

### Step 4: Evaluate the Integral
To solve \( \int t\sqrt{t^2 + 1} \, dt \), we use substitution:
Let \( u = t^2 + 1 \), then \( du = 2t \, dt \), so \( t \, dt = \frac{1}{2} du \).

The integral becomes:
\[
M = 3 \int_{1}^{5} \frac{1}{2} \sqrt{u} \, du = \frac{3}{2} \int_{1}^{5} u^{1/2} \, du
\]
Integrating:
\[
M = \frac{3}{2} \left[\frac{2}{3} u^{3/2} \right]_{1}^{5} = \left[u^{3/2} \right]_{1}^{5} = \left[ (5)^{3/2} - (1)^{3/2} \right]
\]
\[
M = \left[ 5\sqrt{5} - 1 \right]
\]

### Final Answer
The mass of the wire is \( 5\sqrt{5} - 1 \).

Would you like further details or have any questions?

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Here are 5 related questions for further practice:
1. How would the mass change if the density function were \( \delta(t) = 2t^2 \)?
2. What is the total length of the wire along the given curve?
3. How would you compute the center of mass of the wire?
4. What if the curve were parameterized differently, such as \( \mathbf{r}(t) = \cos(t)\mathbf{i} + \sin(t)\mathbf{j} + t\mathbf{k} \)?
5. How would the problem change if the limits of \( t \) were \( 0 \leq t \leq 3 \)?

**Tip:** When solving integrals involving square roots, substitution is often a helpful technique to simplify the expression.

Learn how to calculate the mass of a wire along a parametric curve using integral calculus and vector calculus. Follow step-by-step instructions with detailed formulas and integration techniques. Suitable for advanced college-level mathematics.

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