Consider a curve lying on the cylinder x^2 + y^2 = 1 and given by the vector function r(t) = (cost,sin t, t^2), for t ≥ 0.
Find equations of normal planes at the points (1, 0, 0) and (1, 0, 4π^2)
Hi, I don't really know how to start this. I know the formula for a plane. a(x-xo)+b(y-yo)+c(z-zo)=0 I guess I need help finding the normal vectors
To find the equations of the normal planes at the given points, we need to first find the normal vectors to the curve at those points.
The normal vector to a curve at a particular point is obtained by taking the derivative of the vector function with respect to the parameter t, and evaluating it at the specific value of t corresponding to the point.
Let's start by finding the derivative of the vector function r(t):
r'(t) = (-sin(t), cos(t), 2t)
Now, we can find the normal vector at each point:
For the point (1, 0, 0), we need to evaluate r'(t) at t = 0:
r'(0) = (-sin(0), cos(0), 2*0) = (0, 1, 0)
So, the normal vector to the curve at the point (1, 0, 0) is (0, 1, 0).
Similarly, for the point (1, 0, 4π^2), we need to evaluate r'(t) at t = 2π:
r'(2π) = (-sin(2π), cos(2π), 2*2π) = (0, -1, 4π)
So, the normal vector to the curve at the point (1, 0, 4π^2) is (0, -1, 4π).
Now that we have the normal vectors, we can proceed to find the equations of the normal planes at each point using the formula you mentioned:
For the point (1, 0, 0) with the normal vector (0, 1, 0), the equation of the normal plane is:
0(x - 1) + 1(y - 0) + 0(z - 0) = 0
Simplifying:
y = 0
So, the equation of the normal plane at (1, 0, 0) is y = 0.
For the point (1, 0, 4π^2) with the normal vector (0, -1, 4π), the equation of the normal plane is:
0(x - 1) + (-1)(y - 0) + 4π(z - 4π^2) = 0
Simplifying:
-y + 4π^2z - 16π^3 = 0
So, the equation of the normal plane at (1, 0, 4π^2) is -y + 4π^2z - 16π^3 = 0.