Question

Help finding the equation of normal and osculating plane given x = 2sin3t, y = t, z = 2cos3t and given point (0,pi, -2)

Answer 1

The normal plane at P has normal vector \[r \prime(\frac{ \pi }{ 2 })\]and passes through the point\[(0, \pi,-2)\]
We're given \[r(t)=(2\sin 3t)i +tj +(2\cos 3t)k\] Thus
\[r \prime(t)=(6\cos 3t)i +j -(6\sin 3t)k\]Now
\[r \prime(\frac{ \pi }{ 2 })=j +6k\]In component form, this vector is\[[0, 1, 6]\]
∴ the equation of the normal plane is \[0(x -0)+1(y -\pi)+6(z +2)=0\]OR
\[y -\pi +6z +12=0\]OR
\[6z =-y +\pi -12\]OR
\[z =-\frac{ 1 }{ 6 }y +\frac{ \pi }{ 6 }-2\]

Answer 2

You ponder that while I type a detailed explanation for the equation of an osculating plane.

Answer 3

Thanks for your help, I was finally able to grasp the concept with your assistance!

Answer 4

Now the osculating plane through the given point (0, π, -2) contains the unit tangent vector \[T(t) =\frac{ r \prime(t) }{\left| r \prime(t) \right| }\]and the principal unit normal vector\[N(t)=\frac{ T \prime(t) }{ \left| T \prime(t) \right| }\]hence it's normal vector is the cross product of the unit tangent vector and the principal unit normal vector, and is given by the binormal vector \[B(t)\] where \[B(t)\] as mentioned is the cross product of the unit tangent vector and the principal unit normal vector.

Before we continue, note that \[\left| r \prime(t) \right|and \left| T \prime(t) \right|\]refer to the magnitudes of those vectors. From my previous solution,\[r \prime(t)=(6\cos 3t)i +j -(6\sin 3t)k\]
\[\left| r \prime(t) \right|=\sqrt{(6\cos 3t)^{2}+1^{2}+(-6\sin 3t)^{2}}\]
\[\left| r \prime(t) \right|=\sqrt{36\cos ^{2}3t +1+36\sin ^{2}3t}\]
\[\left| r \prime(t) \right|=\sqrt{37}\]Thus\[T(t)=\frac{ \sqrt{37} }{ 37 }[(6\cos 3t)i +j -(6\sin 3t)k]\]
Now similarly\[T \prime(t)=\frac{ \sqrt{37} }{ 37 }[(-18\sin 3t)i -(18\cos 3t)k]\]
\[\left| T \prime(t) \right|=\sqrt{\frac{ 37 }{ 1369 }(-18\sin 3t)^{2}+(-18\cos 3t)^{2}}\]
\[\left| T \prime(t) \right|=\sqrt{\frac{ 37 }{ 1369 }(324)}\]
Tell me if you know how to finish the rest on your own. If not then I will help you and complete it.

Answer 5

I finished it. Thanks again!

Answer 6

You're welcome. Sorry it seemed like it was taking long. It's just that I like to take my time and make sure I haven't made careless errors, which doesn't really help the person I'm teaching. Although sometimes I inadvertently make them anyway. lol