emancipezN

2021-01-08

Investigation Consider the helix represented by the vector-valued function$r(t)=\text{}\text{}2\text{}\mathrm{cos}\text{}t,\text{}2\text{}\mathrm{sin}\text{}t,\text{}t\text{}$ (a) Write the length of the arc son the helix as a function of t by evaluating the integral$s=\text{}{\int}_{0}^{t}\text{}\sqrt{[{x}^{\prime}(u){]}^{2}\text{}+\text{}[{y}^{\prime}(u){]}^{2}\text{}+\text{}[{z}^{\prime}(u){]}^{2}\text{}du}$

crocolylec

Skilled2021-01-09Added 100 answers

To calculate: The length of the arc s on the helix as a function of tThe length of the curve is $\underset{\u2015}{s=\text{}\sqrt{5t}}.$ Used formula: $s=\text{}{\int}_{0}^{t}\text{}\sqrt{[{x}^{\prime}(t){]}^{2}\text{}+\text{}[{y}^{\prime}(t){]}^{2}\text{}+\text{}[{z}^{\prime}(t){]}^{2}\text{}dt}$ Calculation:The helix path is,$r(t)=\text{}\text{}2\text{}\mathrm{cos}\text{}t,\text{}2\text{}\mathrm{sin}\text{}t,\text{}t\text{}$ On differentiating this vector value function,${r}^{\prime}(t)=\text{}\text{}-2\text{}\mathrm{sin}\text{}t,\text{}2\text{}\mathrm{cos}\text{}t,\text{}1\text{}$ Calculate the length of the line segment for the given interval as$s=\text{}{\int}_{0}^{t}\text{}\sqrt{[{x}^{\prime}(t){]}^{2}\text{}+\text{}[{y}^{\prime}(t){]}^{2}\text{}+\text{}[{z}^{\prime}(t){]}^{2}\text{}dt}$

$=\text{}{\int}_{0}^{t}\text{}\sqrt{(-2\text{}\mathrm{sin}\text{}t{)}^{2}\text{}+\text{}(2\text{}\mathrm{cos}\text{}t{)}^{2}\text{}+\text{}(1{)}^{2}\text{}dt}$

$=\text{}{\int}_{0}^{t}\text{}\sqrt{5dt}$

$=\sqrt{5t}$ Thus, the arc length is $s=\sqrt{5t}.$

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