Question

Find the point on the curve
r(t) = (5 sin t)i + (5 cost)j + 12t k
at a distance 267 units along the curve from the origin in the
direction of increasing arc length.​

Answer 1

The point on the curve r(t) = (5 sin t)i + (5 cost)j + 12t k

at a distance of 267 units along the curve from the origin in the

direction of increasing arc length is (0, 0, 20.53)

Given that;

r(t) = (5 sin t)i + (5 cost)j + 12t k

The distance along the curve from the origin = 267 units

To find;

The point on the curve r(t) = (5 sin t)i + (5 cost)j + 12t k

at a distance of 267 units along the curve from the origin in the

direction of increasing arc length.​

Solution;

We have,

r(t) = (5 sin t)i + (5 cost)j + 12tk

differentiating the above equation we get,

dr(t)/dt = 5cost i - 5sin tj + 12k and its length would be,

$\sqrt{25cos^{2} + 25sin^{2} + 144}$ = 13 . So arc length from t = 0 to t = u is,

∫ 13 dt  = 13u

The asked for point would be when t = 20.53

And so it is (0, 0, 20.53)

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