Blog: Find the arc length of the cuve r(t)=(cos^3 t) i + (sin^3 t) j + 2k , 0=

Friday, March 7, 2014

Find the arc length of the cuve r(t)=(cos^3 t) i + (sin^3 t) j + 2k , 0=

You need to determine the length of the vector function $\displaystyle{\overline{{r}}}{\left({t}\right)}=$ $\displaystyle\lt{{\cos}}^{{3}}{t},{{\sin}}^{{3}}{t},{2}\gt,$ over interval $\displaystyle{\left[{0},\frac{\pi}{{2}}\right]},$ hence you need to find the magnitude of the tangent vector such that:

$\displaystyle{\left|{\overline{{r}}}'{\left({t}\right)}\right|}=\sqrt{{{{\left({3}{{\cos}}^{{2}}{t}\cdot{\left(-{\sin{{t}}}\right)}\right)}}^{{2}}+{{\left({3}{{\sin}}^{{2}}{t}\cdot{\cos{{t}}}\right)}}^{{2}}}}$

$\displaystyle{\left|{\overline{{r}}}'{\left({t}\right)}\right|}=\sqrt{{{9}{{\cos}}^{{4}}{t}\cdot{{\sin}}^{{2}}{t}+{9}{{\sin}}^{{4}}{t}\cdot{{\cos}}^{{2}}{t}}}$

$\displaystyle{\left|{\overline{{r}}}'{\left({t}\right)}\right|}=\sqrt{{{9}{{\cos}}^{{2}}{t}\cdot{{\sin}}^{{2}}{t}{\left({{\cos}}^{{2}}{t}+{{\sin}}^{{2}}{t}\right)}}}$

You need to remember that $\displaystyle{{\cos}}^{{2}}{t}+{{\sin}}^{{2}}{t}={1}$ such that:

$\displaystyle{\left|{\overline{{r}}}'{\left({t}\right)}\right|}=\sqrt{{{9}{{\cos}}^{{2}}{t}\cdot{{\sin}}^{{2}}{t}}}$

$\displaystyle{\left|{\overline{{r}}}'{\left({t}\right)}\right|}={3}{\sin{{t}}}\cdot{\cos{{t}}}$

You need to remember the formula of arc length of a curve such that:

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left|{\overline{{r}}}'{\left({t}\right)}\right|}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{3}{\sin{{t}}}\cdot{\cos{{t}}}{\left.{d}{t}\right.}$

You need to multiply and divide the integrand $\displaystyle{\sin{{t}}}\cdot{\cos{{t}}}$ by 2 to obtain the formula of sine of double angle such that:

$\displaystyle{L}={\left(\frac{{3}}{{2}}\right)}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{2}{\sin{{t}}}\cdot{\cos{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\left(\frac{{3}}{{2}}\right)}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle{L}=-{\left(\frac{{3}}{{2}}\right)}\frac{{{\cos{{\left({2}{t}\right)}}}}}{{2}}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle{L}=-{\left(\frac{{3}}{{4}}\right)}{\left({\cos{\pi}}-{\cos{{0}}}\right)}$

$\displaystyle{L}=-{\left(\frac{{3}}{{4}}\right)}{\left(-{1}-{1}\right)}$

$\displaystyle{L}=\frac{{3}}{{2}}$

Hence, evaluating the arc length of the curve given in vectorial form yields $\displaystyle{L}=\frac{{3}}{{2}}$ .

Blog

Friday, March 7, 2014

Find the arc length of the cuve r(t)=(cos^3 t) i + (sin^3 t) j + 2k , 0=

No comments:

Post a Comment

How far is Iago justified in hating Othello?