Show that the projection into the xy-plane of the curve of intersection of the parabolic cylinder z=1−2y^2 and the paraboloid z=x^2+y^2 is an ellipse.
Find a vector-parametric equation r→1(t)=⟨x(t),y(t),z(t)⟩ for the ellipse in the xy-plane.
Shadow: r→1(t)=____ for 0<t<pi/2
well, the intersection is where
1-2y^2 = x^2+y^2
x^2 + 3y^2 = 1
or
x^2 + y^2/(1/3) = 1
x = r cost
y = 1/√3 r sint
To show that the projection into the xy-plane of the curve of intersection of the given surfaces is an ellipse, we need to find the equation of this curve in the form x = f(t) and y = g(t), where t is a parameter.
1. Start by setting the equations of the parabolic cylinder and the paraboloid equal to each other:
1 - 2y^2 = x^2 + y^2
2. Combine like terms on the right side:
0 = x^2 + y^2 - 2y^2 - 1
3. Simplify the equation:
0 = x^2 - y^2 - 1
4. Rewrite the equation in terms of the parameter t by substituting x = x(t) and y = y(t):
0 = x(t)^2 - y(t)^2 - 1
5. To find x(t) and y(t), we can solve this equation for x(t) and y(t) separately:
x(t)^2 = y(t)^2 + 1 --> x(t) = ±√(y(t)^2 + 1)
6. Since we are looking for a curve in the xy-plane, we can ignore the positive square root and take the negative square root:
x(t) = -√(y(t)^2 + 1)
7. We can choose y(t) as the parameter itself, so let y(t) = t.
8. Substitute y(t) = t into the equation for x(t):
x(t) = -√(t^2 + 1)
9. Therefore, the vector-parametric equation r→1(t) for the ellipse in the xy-plane is:
r→1(t) = ⟨-√(t^2 + 1), t, 0⟩ for 0 < t < π/2
Note: Since we are projecting into the xy-plane, the z-component is always zero.
To show that the projection into the xy-plane of the curve of intersection is an ellipse, we need to eliminate the parameter z from the equations of the parabolic cylinder and the paraboloid.
1. Equation of the parabolic cylinder: z = 1 - 2y^2
2. Equation of the paraboloid: z = x^2 + y^2
Setting the two equations equal to each other, we get:
1 - 2y^2 = x^2 + y^2
Rearranging the equation, we obtain:
x^2 + 2y^2 + y^2 - 1 = 0
Combining like terms, we have:
x^2 + 3y^2 - 1 = 0
This equation represents an ellipse in the xy-plane since the coefficients of x^2 and y^2 have opposite signs.
Now, let's find a vector-parametric equation for the ellipse in the xy-plane.
We can parameterize the ellipse by using trigonometric functions. Let's assume:
x(t) = cos(t)
y(t) = (1/√3)sin(t)
Substituting these parameterizations into the equation of the ellipse, we have:
(cos(t))^2 + 3(sin(t))^2 - 1 = 0
Simplifying the equation, we get:
cos^2(t) + 3sin^2(t) = 1
Using the identity cos^2(t) + sin^2(t) = 1, we can rewrite the equation as:
2sin^2(t) + sin^2(t) = 1
Simplifying again, we obtain:
sin^2(t) = 1/3
Taking the square root of both sides, we have:
sin(t) = ± √(1/3)
Since the parameter t lies between 0 and π/2, we can use the positive square root:
sin(t) = √(1/3)
Plugging this value back into our parameterization, we get:
x(t) = cos(t)
y(t) = (1/√3)√(1/3) = 1/3
Therefore, the vector-parametric equation for the ellipse in the xy-plane is:
r→1(t) = <cos(t), (1/3)sin(t), 0>
for 0 < t < π/2.