Chapter 9: Q 9. (page 775)

Find the arc lengths of the curves defined by the parametric equations on the specified intervals.
$x = t^{2}$ , $y = t^{3}$ , $t \in [- 2, 2]$

Short Answer

Expert verified

The required length of the curve is $\begin{array}{rcl} \frac{2}{27} \{40^{\frac{3}{2}} - 4^{\frac{3}{2}}\} \end{array}$ .

Step by step solution

Step 1. Given information.

The given parametric equations are $x = t^{2}$ , $y = t^{3}$ .

Step 2. Find the derivative of the parametric equations.

$\begin{array}{rcl} x & = & t^{2} \\ f^{'} (t) & = & 2 t \end{array}$ and $\begin{array}{rcl} y & = & t^{3} \\ g' (t) & = & 3 t^{2} \end{array}$

Step 3. Substitute the value of f'(t) and g'(t) in the formula.

The length of the curve is:

$\int_{a}^{b} \sqrt{{(f' (t))}^{2} + {(g' (t))}^{2}} d t$ , where $[a, b] = [- 2, 2]$ .

So, the length of the curve is:

localid="1652254615405" $\begin{array}{rcl} \int_{a}^{b} \sqrt{{(f' (t))}^{2} + {(g' (t))}^{2}} d t & = & \int_{- 2}^{2} \sqrt{{(2 t)}^{2} + {(3 t^{2})}^{2}} d t \\ = & \int_{- 2}^{2} \sqrt{4 t^{2} + 9 t^{4}} d t \\ = & \int_{- 2}^{2} \sqrt{t^{2} (4 + 9 t^{2})} d t . . . . . . . . . . . . . . . . . (1) \end{array}$

Step 4. Now solve the indefinite integral.

Substitute $4 + 9 t^{2} = z$ .

$\begin{array}{rcl} 18 t d t & = & d z \\ d t & = & \frac{d z}{18 t} \end{array}$

$\begin{array}{rcl} \int \sqrt{t^{2} (4 + 9 t^{2})} d t & = & \int t \sqrt{(4 + 9 t^{2})} d t \\ = & \int t \sqrt{z} \frac{d z}{18 t} \\ = & \int \sqrt{z} \frac{d z}{18} \\ = & \frac{1}{18} \int \sqrt{z} d z \\ = & \frac{1}{18} \cdot \frac{2}{3} z^{\frac{3}{2}} \\ = & \frac{1}{27} {(4 + 9 t^{2})}^{\frac{3}{2}} \end{array}$

Step 5. Apply limits to solve definite integral.

$\begin{array}{rcl} \int_{- 2}^{2} \sqrt{t^{2} (4 + 9 t^{2})} d t & = & 2 \int_{0}^{2} \sqrt{t^{2} (4 + 9 t^{2})} d t \\ = & 2 {[\frac{1}{27} {(4 + 9 t^{2})}^{\frac{3}{2}}]}_{0}^{2} \\ = & 2 [\frac{1}{27} {(4 + 9 {(2)}^{2})}^{\frac{3}{2}}] - 2 [\frac{1}{27} {(4 + 9 {(0)}^{2})}^{\frac{3}{2}}] \\ = & 2 [\frac{1}{27} {(40)}^{\frac{3}{2}}] - 2 [\frac{1}{27} {(4)}^{\frac{3}{2}}] \\ = & \frac{2}{27} \{40^{\frac{3}{2}} - 4^{\frac{3}{2}}\} \end{array}$

Step 6. Simplified answer.

Hence, the required arc length is $\begin{array}{rcl} \frac{2}{27} \{40^{\frac{3}{2}} - 4^{\frac{3}{2}}\} \end{array}$ .

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Find the arc lengths of the curves defined by the parametric equations on the specified intervals.
$x = t^{2}$ , $y = t^{3}$ , $t \in [- 2, 2]$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the derivative of the parametric equations.

Step 3. Substitute the value of f'(t) and g'(t) in the formula.

Step 4. Now solve the indefinite integral.

Step 5. Apply limits to solve definite integral.

Step 6. Simplified answer.

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Find the arc lengths of the curves defined by the parametric equations on the specified intervals.x=t2,y=t3,t∈-2,2

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the derivative of the parametric equations.

Step 3. Substitute the value of f'(t) and g'(t) in the formula.

Step 4. Now solve the indefinite integral.

Step 5. Apply limits to solve definite integral.

Step 6. Simplified answer.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Geometry

Theoretical and Mathematical Physics

Statistics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Find the arc lengths of the curves defined by the parametric equations on the specified intervals.
$x = t^{2}$ , $y = t^{3}$ , $t \in [- 2, 2]$