Answer:
the answer is incomplete, below is the complete question
"Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 3ti + (1 - 4t)j + (1 + 2t)k r(t(s)) ="
answer
[tex]r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k[/tex]
Step-by-step explanation:
The step by step procedure is to first determine the differentiate the given vector function
r(t) = 3ti + (1 - 4t)j + (1 + 2t)k
[tex]\frac{d(r(t) = 3ti + (1 - 4t)j + (1 + 2t)k)}{dt} \\r'(t)=3i-4j+2k\\[/tex]
since s(t) is the arc length for r(t), which is define as
[tex]s(t)=\int\limits^t_0 {||r'(t)||} \, dt[/tex]
if we substitute the value of r'(t) we arrive at
[tex]s(t)=\int\limits^t_0 {||r'(t)||} \, dt\\s(t)=\int\limits^t_0 {\sqrt{3^{2} +4^{2}+2^{2}} \, dt\\s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\[/tex]
[tex]s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\\\s(t)=\sqrt{29} t\\hence \\t(s)=\frac{s}{\sqrt{29} }[/tex]
substituting the value of t in to the given vector equation we have
[tex]r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k[/tex]
