Find the length of the curve. r→(t)=&lt;8t,t2,112t3&gt;,0≤t≤1

Question

Find the length of the curve.   r→(t)=&amp;lt;8t,t2,112t3&amp;gt;,0≤t≤1

esfloravaou · Accepted Answer

r→′(t)=&amp;lt;8,2t,312t2&amp;gt;r→′(t)=&amp;lt;8,2t,14t2&amp;gt;||r→′(t)||=(8)2+(2t)2+(14t2)2⇒||r→′(t)||=64+4t2+116t4⇒||r→′(t)||=116((64⋅16)+64t2+t4)⇒||r→′(t)||=141024+64t2+t4⇒||r→′(t)||=14322+(2⋅32)t2+t4⇒||r→′(t)||=14(32+t2)2⇒||r→′(t)||=14|32+t2|Now, t2 is always positive. So, (32+t2) is always positive.

Cheryl King · Accepted Answer

Where is the second part of solution?

RizerMix · Accepted Answer

we know that, the length of the curve between t = 0 and t = 1 can be computed as L=∫t=0t=1||r→(t)||dt⇒L=∫t=0t=114(32+t2)dt⇒L=14∫t=0t=1(32+t2)dt⇒L=14[32+t33]t=0t=1⇒L=14[(32+13)−(0+0)]⇒L=14(32+13)⇒L=14∗973⇒L=9712

Find the length of the curve. r→(t)=<8t,t2,112t3>,0≤t≤1

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