To find the space curve of intersection, we need to solve for a common set of values for x, y, and z that satisfy both equations simultaneous. Therefore, we have:
3x^2 + y^2 + 1 = z (Equation 1)
x^2 + 3y^2 + z = 5 (Equation 2)
We can start by isolating z in Equation 1:
z = 3x^2 + y^2 + 1 (Equation 3)
Next, we can substitute Equation 3 into Equation 2 for z:
x^2 + 3y^2 + (3x^2 + y^2 + 1) = 5
Simplifying, we get:
4x^2 + 4y^2 = 4
x^2 + y^2 = 1
This is the equation of a cylinder with radius 1 centered at the origin, which is a circle in the xy-plane. To parameterize this circle, we can use:
x = cos(t)
y = sin(t)
Substituting these into Equation 3, we get:
z = 3cos^2(t) + sin^2(t) + 1
z = 2cos^2(t) + 1
Therefore, the space curve of intersection is:
r(t) = cos(t) i + sin(t) j + (2cos^2(t) + 1) k
Find the space curve, vector function r(t), where the following surfaces intersect:
3x^2+y^2+1=z, x^2+3y^2+z=5
3 answers
pretty good, except
3cos^2(t) + sin^2(t) + 1 = 2cos^2(t)+2
3cos^2(t) + sin^2(t) + 1 = 2cos^2(t)+2
You are correct! I apologize for the mistake.
The corrected version is:
r(t) = cos(t) i + sin(t) j + (2cos^2(t) + 1) k
The corrected version is:
r(t) = cos(t) i + sin(t) j + (2cos^2(t) + 1) k