Find the curvature of the curve r(t).
-r(t) = (10 + 6 cos 8t) i - (4 + 6 sin 8t)j + 3k
A) $\kappa = \frac { 1 } { 6 }$
B) $\kappa = \frac { 4 } { 3 }$
C) $\kappa = 6$
D) $\kappa = \frac { 1 } { 36 }$

Question

Find the curvature of the curve r(t).
-r(t) = (10 + 6 cos 8t) i - (4 + 6 sin 8t)j + 3k 
A) $\kappa = \frac { 1 } { 6 }$
B) $\kappa = \frac { 4 } { 3 }$
C) $\kappa = 6$
D) $\kappa = \frac { 1 } { 36 }$

Quizplus · Accepted Answer

To find the curvature of the curve, we first need to find the unit tangent and unit normal vectors.

r(t) = (10 + 6 cos 8t) i - (4 + 6 sin 8t)j + 3k

Taking the first derivative, we get:

r'(t) = -48 sin 8t i - 48 cos 8t j

Taking the second derivative, we get:

r''(t) = -384 cos 8t i + 384 sin 8t j

Now we can find the magnitude of r'(t):

|r'(t)| = sqrt((-48 sin 8t)^2 + (-48 cos 8t)^2) = 48

So the unit tangent vector is:

T(t) = r'(t)/|r'(t)| = (-sin 8t)i - (cos 8t)j

To find the unit normal vector, we take the derivative of T(t):

T'(t) = -8 cos 8t i + 8 sin 8t j

And then we find its magnitude:

|T'(t)| = 8sqrt(cos^2(8t) + sin^2(8t)) = 8

So the unit normal vector is:

N(t) = T'(t)/|T'(t)| = -cos 8t i + sin 8t j

Finally, we can find the curvature:

κ = |T'(t)|/|r'(t)| = 8/48 = 1/6

Therefore, the answer is A, with κ = 1/6.

Find the Curvature of the Curve R(t) κ=16\kappa = \frac { 1 } { 6 }κ=61​

Find the Curvature of the Curve R(t) $\kappa = \frac { 1 } { 6 }$