The plane determined by the unit normal and binormal vectors N and B at a point

Question

The plane determined by the unit normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P, and the plane determined by the unit vectors T and N is called the osculating plane of C at P. Consider the curve C described by the position vector r(t) = t^3 i + 3tj + t^4 k. When is the normal plane on the curve parallel to the plane 6x + 6y - 8z = 1? Call this time t*. Determine the curvature of the path at time t*. What are the coordinates of the center of the osculating circle to the curve C at time t*? What is the standard equation of the osculating plane for path C at time t* ?

Explanation / Answer

n=(6,6,-8) is a normal vecttor to the plane . We have : r(t)=(3t^2,3,4t^3) is a tangent line director vector is parallel to the plane if it's normal to n, thus : n.r'(t)=0, thus :

6(3t^2)+6(3)-8(4t^3)=0

18t^2+18-32t^3=0 will be the equation

To find t we will apply desmos graph    we get

t= 1.1

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