(1 point)
Find the length of the parametric curve
$x = 1 + 27 t^{2}, y = 6 + 18 t^{3}, 0 \leq t \leq 5 .$

Answer

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Answer

So, the length of the parametric curve is $3024$ .

Steps

Step 1 ：First, we need to find $\frac{d x}{d t}$ and $\frac{d y}{d t}$ .

Step 2 ：We have $\frac{d x}{d t} = \frac{d (1 + 27 t^{2})}{d t} = 54 t$ and $\frac{d y}{d t} = \frac{d (6 + 18 t^{3})}{d t} = 54 t^{2}$ .

Step 3 ：Then, we substitute these into the formula for the length of a parametric curve: $L = \int_{a}^{b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ .

Step 4 ：This gives us $L = \int_{0}^{5} \sqrt{(54 t)^{2} + (54 t^{2})^{2}} d t = \int_{0}^{5} \sqrt{2916 t^{2} + 2916 t^{4}} d t = \int_{0}^{5} \sqrt{2916 (t^{2} + t^{4})} d t = \int_{0}^{5} 54 t \sqrt{1 + t^{2}} d t$ .

Step 5 ：This integral is not easy to solve directly, so we use a substitution: let $u = t^{2} + 1$ , then $d u = 2 t d t$ , and $d t = \frac{d u}{2 t}$ .

Step 6 ：Substituting these into the integral, we get $L = \int_{1}^{26} 27 \sqrt{u} d u = [18 u^{3 / 2}]_{1}^{26} = 18 * 26^{3 / 2} - 18 * 1^{3 / 2} = 18 * 169 - 18 = 3024$ .

Step 7 ：So, the length of the parametric curve is $3024$ .

(1 point)Find the length of the parametric curvex=1+27t2,y=6+18t3,0≤t≤5.

Answer

Popular Problems

(1 point)
Find the length of the parametric curve
$x = 1 + 27 t^{2}, y = 6 + 18 t^{3}, 0 \leq t \leq 5 .$