Given the vector&nbsp;r(t)=cos⁡T,sin⁡T,ln⁡(cos⁡T)&nbsp;and point (1, 0, 0) find vectors T, N and B at that point.Vector T is the unit tangent vector, so the derivative r(t) is needed.Vector N is the normal unit vector, and the equation for it uses the derivative of T(t).&nbsp;The B vector is the binormal vector, which is a crossproduct of T and N.

Question

Given the vector&amp;nbsp;r(t)=cos⁡T,sin⁡T,ln⁡(cos⁡T)&amp;nbsp;and point (1, 0, 0) find vectors T, N and B at that point.Vector T is the unit tangent vector, so the derivative r(t) is needed.Vector N is the normal unit vector, and the equation for it uses the derivative of T(t).&amp;nbsp;The B vector is the binormal vector, which is a crossproduct of T and N.

Eliza Beth13 · Accepted Answer

To find the vectors T, N, and B at the point (1, 0, 0) for the given vector function r(t)=cos(t),sin(t),ln(cos(t)), we need to calculate the derivatives and perform some vector operations.First, let&#039;s find the unit tangent vector T. The unit tangent vector is the derivative of r(t), normalized to have a magnitude of 1:T(t)=dr/dt‖dr/dt‖Taking the derivative of r(t), we have:r′(t)=(−sin(t),cos(t),−sin(t)cos(t))Simplifying, we get:r′(t)=(−sin(t),cos(t),−tan(t))To find T(t), we need to normalize r&#039;(t):T(t)=r′(t)‖r′(t)‖Calculating the magnitude of r&#039;(t):‖r′(t)‖=sin2(t)+cos2(t)+tan2(t)Simplifying further:‖r′(t)‖=1+tan2(t)=sec2(t)=sec(t)Now, we can find T(t):T(t)=(−sin(t)sec(t),cos(t)sec(t),−tan(t)sec(t))Simplifying, we have:T(t)=(−sin(t)cos(t),cos2(t),−sin(t))To find the normal unit vector N, we differentiate T(t) with respect to t:N(t)=dT/dt‖dT/dt‖Differentiating T(t), we have:dTdt=(cos2(t),−2sin(t)cos(t),−cos(t))Calculating the magnitude of dT/dt:‖dT/dt‖=cos4(t)+4sin2(t)cos2(t)+cos2(t)Simplifying:‖dT/dt‖=cos2(t)(cos2(t)+4sin2(t)+1)=cos2(t)(cos2(t)+sin2(t)+1)Since cos2(t)+sin2(t)=1, we can further simplify:‖dT/dt‖=cos2(t)·2=2cos(t)Now, we can find N(t):N(t)=(cos2(t)2cos(t),−2sin(t)cos(t)2cos(t),−cos(t)2cos(t))Simplifying, we have:N(t)=(cos(t)2,−2sin(t)2,−cos(t)2)Simplifying further:N(t)=(cos(t)2,−sin(t),−cos(t)2)Finally, to find the binormal vector B, we can take the cross product of T(t) and N(t):B(t)=T(t)×N(t)Calculating the cross product, we have:B(t)=((−2cos(t)sin(t))/2,(cos2(t))/2−(cos2(t))/2,(sin(t))/2−(−sin(t))/2)Simplifying, we get:B(t)=(−cos(t)sin(t),0,sin(t))Now, we can evaluate these vectors at the point (1, 0, 0) by substituting t=0 into the expressions for T(t), N(t), and B(t):T=(−sin(0)cos(0),cos2(0),−sin(0))=(0,1,0)N=(cos(0)2,−sin(0),−cos(0)2})=(12,0,−12)B=(−cos(0)sin(0),0,sin(0))=(0,0,0)Therefore, at the point (1, 0, 0), the vectors T, N, and B are (0, 1, 0), (12,0,−12), and (0, 0, 0), respectively.

Given the vector r(t)=cos⁡T,sin

Answered question

Answer & Explanation

New Questions in Differential Calculus