Explained: "Find the length of the curve. r(t)=8i+t2j+t3k, 0≤t≤1"

Idilwsiw2

Answered question

2021-11-13

Find the length of the curve.

r (t) = 8 i + t^{2} j + t^{3} k, 0 \leq t \leq 1

breisgaoyz

Beginner2021-11-14Added 14 answers

Given:
The curve

r (t) = 8 i + t^{2} j + t^{3} k

for

0 \leq t \leq 1

To find: The length of the curve.
Answer: The arc length is given by

| | r (t) | | = \int_{a}^{b} \sqrt{{(x^{'})}^{2} + {(y^{'})}^{2} + {(z^{'})}^{2}} d t

where

r (t) = ξ + y j + z k

obtain the value as

x^{'} = \frac{d}{d t} (8) = 0

y^{'} = \frac{d}{d t} (t^{2}) = 2 t

z^{'} = \frac{d}{d t} (t^{3}) = 3 t^{2}

Substitute the values in the formula.
The length for the interval

0 \leq t \leq 1

becomes

| | r (t) | | = \int_{0}^{1} \sqrt{{(0)}^{2} + {(2 t)}^{2} + {(3 t^{2})}^{2}} d t

= \int_{0}^{1} \sqrt{4 t^{2} + 9 t^{4}} d t

\int_{0}^{1} t \sqrt{4 + 9 t^{2}} d t

Take

s = 4 + 9 t^{2}

, differentiating with respect to t gives

d s = 18 t d t

t d t = \frac{1}{18} d s

t = 0, s = 4 + 9 {(0)}^{2} = 4

t = 1, s = 4 + 9 {(1)}^{2} = 13

Thus, the integral with the above substion becomes

| | r (t) | | = \int_{0}^{1} t \sqrt{4 + 9 t^{2}} d t

= \frac{1}{18} \int_{4}^{13} \sqrt{s} d s

= \frac{1}{18} {[\frac{2}{3} s^{\frac{3}{2}}]}_{4}^{13}

=

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