Question

Consider the following. x=sin(4t),y=−cos(4t),z=8t;(0,1,2π) Find the equation of the normal plane of the curve at the givent. Find the equation of the osculating plane of the curve at the given point.

Answer 1

To find the equation of the normal plane at the given point (0,1,2π), we calculate the derivative and substitute the values into the equation of a plane. The normal plane equation is 4x + 8z - 16 = 0. To find the equation of the osculating plane, we calculate the second derivative and substitute the values into the equation of a plane. The osculating plane equation is -16y = 0.

Step-by-step explanation:

To find the equation of the normal plane of the curve at the given point (0,1,2π), we first need to calculate the derivative of the given curve. The derivative is obtained by taking the derivative of each component of the curve with respect to t. So, we have dx/dt = 4cos(4t), dy/dt = 4sin(4t), and dz/dt = 8.

Next, we substitute the given point (0,1,2π) into the derivative. So, dx/dt = 4cos(4(2π)) = 4cos(8π) = 4cos(0) = 4, dy/dt = 4sin(4(2π)) = 4sin(8π) = 4sin(0) = 0, and dz/dt = 8.

Now we can use these derivative values to find the equation of the normal plane. The equation of a plane can be written as Ax + By + Cz + D = 0, where the normal vector of the plane is given by n = <4, 0, 8>. Substituting the values of (x, y, z) = (0,1,2π) and (A, B, C) = (4, 0, 8) into the equation, we obtain the equation of the normal plane as 4x + 8z - 16 = 0.

To find the equation of the osculating plane of the curve at the given point, we need to calculate the second derivative of the curve. The second derivative is obtained by taking the derivative of each component of the first derivative with respect to t. So, we have d²x/dt² = -16sin(4t), d²y/dt² = -16cos(4t), and d²z/dt² = 0.

Now, we substitute the given point (0,1,2π) into the second derivative. So, d²x/dt² = -16sin(4(2π)) = -16sin(8π) = -16sin(0) = 0, d²y/dt² = -16cos(4(2π)) = -16cos(8π) = -16cos(0) = -16, and d²z/dt² = 0.

Finally, we can use these second derivative values to find the equation of the osculating plane. The equation of a plane can be written as Ax + By + Cz + D = 0, where the normal vector of the plane is given by n = <0, -16, 0>. Substituting the values of (x, y, z) = (0,1,2π) and (A, B, C) = (0, -16, 0) into the equation, we obtain the equation of the osculating plane as -16y = 0.

Answer 2

To find the equation of the normal plane of the curve at the given point (0,1,2π), we first find the derivatives of the curve and substitute the given value of t. The equation of the normal plane is 4x+(z-2π)(8)=0. To find the equation of the osculating plane, we need to find the second derivatives of the curve and substitute the given value of t. The equation of the osculating plane is -16y+16=0.

Step-by-step explanation:

To find the equation of the normal plane of the curve at the given point (0,1,2π), we first need to find the derivative of the curve. Taking the derivative of x=sin(4t), y=−cos(4t), and z=8t with respect to t, we get dx/dt=4cos(4t), dy/dt=4sin(4t), and dz/dt=8.

Next, we substitute the given value of t=2π into the derivatives to find the slope of the curve at the point. So, dx/dt=4cos(8π)=4 and dy/dt=4sin(8π)=0. Finally, we can use the point-normal form of the plane equation to find the equation of the normal plane, which is given by (x-0)(4)+(y-1)(0)+(z-2π)(8)=0. Simplifying this equation gives 4x+(z-2π)(8)=0.

To find the equation of the osculating plane of the curve at the given point, we need to find the second derivative of the curve. Taking the second derivative of x=sin(4t), y=−cos(4t), and z=8t with respect to t, we get d^2x/dt^2=-16sin(4t), d^2y/dt^2=-16cos(4t), and d^2z/dt^2=0.

Substituting the given value of t=2π into the second derivatives, we get d^2x/dt^2=-16sin(8π)=0 and d^2y/dt^2=-16cos(8π)=-16. The equation of the osculating plane is then given by (x-0)(0)+(y-1)(-16)+(z-2π)(0)=0. Simplifying this equation gives -16y+16=0.