Question

Find the point on the curve r(t) = (5Sint)i + (5Cost)j +12tk
at a distance 26pi units along the curve from the point (0,5,0) inthe direction of increasing arc length.

Answer 1

Answer:

Find the point on the curve r(t) = (5Sint)i + (5Cost)j +12tk

at a distance 26pi units along the curve from the point (0,5,0) inthe direction of increasing arc length.

(My attempt):

T comes to be 2pi and when the integral is done and solved to givea value of 26pi and the position comes to be (0,5,24pi). However,this calculation and answer though correct (according to the backof the book) does not involve the use of the fact that at time t=0,the particle is at (0,5,0).

What for is that information given then?

Step-by-step explanation:

Consider the curve r(t) = (5Sint)i + (5Cost)j + (12t)k

Need to find the point on the given curve at a distance 26π unit along the curve  from the point (0,5,0) inthe direction of increasing arc length.

Length of a smooth curve is $r(t)=x(t)i+y(t)j+z(t)k, \ \ a\leq t\leq b$ that is traced exactly once as t increase from t = a to t = b, is

$L=\int\limits^b_a \sqrt{(\frac{dx}{dt} )^2+(\frac{dy}{dt} )^2+(\frac{dz}{dt} )^2dt}$

For the given curve

x(t) = 5 sin t

y(t) = 5 cos t

z(t) = 12t

When t = 0

x(0) = 5 sin 0

= 0

y(0) = 5 cos 0

= 0

z(0) = 12(0)

=0

So, the point (0, 5, 0) corresponds to t = 0

So let t = t₀ correspond to any point (x, y, z) on the curve at a distance of 26pi  units from the point t = 0 along the increasing arc length

So, the length of curve from the point t = 0 to  t = t₀ is L = 26pi units

Substitute the known value to the arc length formula

$L=\int\limits^b_a \sqrt{(\frac{dx}{dt} )^2+(\frac{dy}{dt} )^2+(\frac{dz}{dt} )^2dt}$

$26\pi=\int\limits^{t_0}\sqrt{(5 \cos t)6+(-5 \sin t)^2+(12)^2dt}\\\\26\pi=\int\limits^{t_0}_0\sqrt{25 \cos ^2t+25 \sin ^2t+144dt} \\\\26\pi=\int\limits^{t_0}_0 \sqrt{25(\cos^2t+ \sin^2t)+144dt}\\\\26\pi=\int\limits^{t_0}_0\sqrt{25(1)+144dt} \\\\26\pi= \int\limits^{t_0}_0\sqrt{169dt} \\\\26\pi= \int\limits^{t_0}_013 dt\\\\26\pi=13\int\limits^{t_0}_0dt\\\\26\pi=13[t]^{t_0}_0\\\\26\pi=13[t_0-0]\\\\26\pi=13t_0\\\\t_0=\frac{26\pi}{13} \\\\t_0=2\pi$

The point corresponding to $t_0 = 2\pi$

when t = 0

$x(2\pi)=5 \sin (2\pi)\\\\=0\\\\y(2\pi)=5 \cos (2\pi)\\\\=5(1)=5\\\\z(2\pi)=12(2\pi)\\\\=24\pi$

Therefore the point corresponding to  $t_0 = 2\pi$ is $(0,5,24\pi)$

Hence, the required point on the given curve at distance 26\pi units along the curve from the point (0,5,0) in  the direction of increasing arc length is $(0,5,24\pi)$