If the curve of intersection of the parabolic cylinder
y = x^2
and the top half of the ellipsoid
x^2 + 5y^2 + 5z^2 = 25.
Then find parametric equations for this curve.
the cylinder has parametric equations
x = t
y = t^2
Intersect that with the ellipsoid and you get
x = t
y = t^2
z = 1/5 √(25-t^2-5t^4)
I got sqrt 5 on the bottom
and you would be correct.
I knew there was something not right.
Well, solving this mathematical conundrum might get a bit twisted, just like a clown in a contortionist act! Let's see if we can untangle this problem.
First, we have the equation of the parabolic cylinder as y = x^2. This means that wherever we plug x into y, it'll be squaring it like a grumpy wizard.
Now, let's focus on the ellipsoid: x^2 + 5y^2 + 5z^2 = 25. It seems like an oblong meatball trying to fit into a round pizza.
To find the curve of intersection, we need to find the values of x, y, and z that satisfy both equations at the same time. It's like searching for a unicorn that poops rainbows!
Substituting y = x^2 into the equation of the ellipsoid, we get: x^2 + 5(x^2) + 5z^2 = 25. This simplifies to 6x^2 + 5z^2 = 25.
Solving for z in terms of x, we have: z^2 = (25 - 6x^2) / 5. So z = ±sqrt((25 - 6x^2) / 5). Looks like our clown is having some fun on a square root trampoline!
Now we have our parametric equations for the curve of intersection:
x = t
y = t^2
z = ±sqrt((25 - 6t^2) / 5)
Just be careful not to trip over those square roots! Happy clowning around with your new parametric equations!
To find the parametric equations for the curve of intersection between the parabolic cylinder and the top half of the ellipsoid, we need to solve the system of equations formed by setting the equations of the two surfaces equal to each other:
y = x^2 ---(1)
x^2 + 5y^2 + 5z^2 = 25 ---(2)
From equation (1), we can substitute the value of y in equation (2):
x^2 + 5(x^2)^2 + 5z^2 = 25 ---(3)
Simplifying equation (3):
6x^4 + 5z^2 = 25 ---(4)
To find the parametric equations, we can express x and z in terms of a parameter t. Let's assume:
x = cos(t)
z = sin(t)
Substituting these values into equation (4), we get:
6(cos(t))^4 + 5(sin(t))^2 = 25
Simplifying further:
6cos^4(t) + 5sin^2(t) = 25
Dividing both sides by 25:
(6/25)cos^4(t) + (5/25)sin^2(t) = 1
Now, we can express cos^2(t) and sin^2(t) in terms of a new parameter u:
cos^2(t) = (1/2)(1 + cos(2u))
sin^2(t) = (1/2)(1 - cos(2u))
Substituting these values into the equation, we get:
(6/25)[(1/2)(1 + cos(2u))]^2 + (5/25)[(1/2)(1 - cos(2u))] = 1
Simplifying further:
(6/25)(1/4)(1 + 2cos(2u) + cos^2(2u)) + (5/25)(1/2)(1 - cos(2u)) = 1
Multiplying everything by 100 to clear fractions:
6(1 + 2cos(2u) + cos^2(2u)) + 10(1 - cos(2u)) = 100
Simplifying:
6 + 12cos(2u) + 6cos^2(2u) + 10 - 10cos(2u) = 100
Combining like terms:
16cos(2u) + 6cos^2(2u) - 84 = 0
Now, we have a quadratic equation in terms of cos(2u). We can solve it to find cos(2u):
6cos^2(2u) + 16cos(2u) - 84 = 0
We can solve this quadratic equation using the quadratic formula:
cos(2u) = [-16 ± √(16^2 - 4(6)(-84))] / 2(6)
Simplifying:
cos(2u) = [-16 ± √(256 + 2016)] / 12
cos(2u) = [-16 ± √(2272)] / 12
cos(2u) = [-16 ± 2√(7)] / 12
cos(2u) = (-4 ± √(7))/3
Now, we have the value of cos(2u). From this, we can find the value of sin(2u) using the identity sin^2(2u) + cos^2(2u) = 1:
sin^2(2u) = 1 - cos^2(2u)
sin^2(2u) = 1 - [(-4 ± √(7))/3]^2
sin^2(2u) = 1 - (16 ± 8√(7) + 7) / 9
sin^2(2u) = (8∓8√(7)) / 9
Now, we can express cos(2u) and sin(2u) in terms of the original parameters x and z:
cos(2u) = 2cos^2(u) - 1 = 2(x^2) - 1
sin(2u) = ± √[1 - cos^2(2u)] = ± √[(8∓8√(7)) / 9]
Therefore, the parametric equations for the curve of intersection are:
x = cos(t)
y = x^2 = (cos(t))^2
z = sin(t)
where cos(t) and sin(t) are given by the values above.