Find the curvature of the space curve.
-r(t) = ti + (sinh t)j + (cosh t)k
A) $\kappa = \sinh ^ { 2 } t$
B) $\kappa = \cosh ^ { 2 } t$
C) $\kappa = \frac { \tanh ^ { 2 } \mathrm { t } } { 2 }$
D) $\kappa = \frac { \operatorname { sech } ^ { 2 } \mathrm { t } } { 2 }$

Question

Find the curvature of the space curve.
-r(t) = ti + (sinh t)j + (cosh t)k 
A) $\kappa = \sinh ^ { 2 } t$
B) $\kappa = \cosh ^ { 2 } t$
C) $\kappa = \frac { \tanh ^ { 2 } \mathrm { t } } { 2 }$
D) $\kappa = \frac { \operatorname { sech } ^ { 2 } \mathrm { t } } { 2 }$

Quizplus · Accepted Answer

First, we need to find the tangent vector and the normal vector of the curve: 
r'(t) = i + cosh(t)j + sinh(t)k 
| r'(t) | = √(1 + cosh^2(t) + sinh^2(t)) = √(cosh^2(t) + sinh^2(t) + 1) = sec(h)t
T(t) = r'(t) / |r'(t)| = (1/sec(h)t)i + (cosh(t)/sec(h)t)j + (sinh(t)/sec(h)t)k
Now, we need to find the derivative of the unit tangent vector: 
T'(t) = (-1/sec^2(t))(-sinh(t))i + (sinh(t)/sec^2(t))j + (cosh(t)/sec^2(t))k
|T'(t)| = √((sinh^2(t)+(-sinh(t))^2+sinh^2(t))/sec^4(t) + cosh^2(t)/sec^4(t)) = √(2(sinh^2(t))/sec^4(t) + cosh^2(t)/sec^4(t)) = √((2sinh^2(t)+cosh^2(t))/sec^4(t))
= √((sinh^2(t)+cosh^2(t))/sec^4(t)) = sech^3(t)/sec^2(t) = sech(t)
N(t) = T'(t)/|T'(t)| = (-sinh(t)/sech(t))i + (cosh(t)/sech(t))j + 0k
Now, we can find the curvature: 
κ = |T'(t)|/|r'(t)| = sech(t)/sec(t) = (1/cos(t))/(1/sec(t)) = cos(t)
Since the parameter t is not mentioned to be in radians or degrees, we can't simplify it to a specific expression. Therefore, the correct answer is D.

Find the Curvature of the Space Curve κ=sinh⁡2t\kappa = \sinh ^ { 2 } tκ=sinh2t

Find the Curvature of the Space Curve $\kappa = \sinh ^ { 2 } t$