Let C be the intersection of x^2+y^2=16 and x+y+z=5. Find the curvature at (0,4,1).

Question

Let C be the intersection of x^2+y^2=16 and x+y+z=5. Find the curvature at (0,4,1).

I don't know how to find the intersection between the given equations.

Answer 1

To find the intersection between the two given equations, we need to solve them simultaneously. Let's start by substituting the value of z from the second equation into the first equation.

From the second equation, we have x + y + z = 5. Rearranging this equation, we get z = 5 - x - y.

Substituting z = 5 - x - y into the first equation x^2 + y^2 = 16, we have:

x^2 + y^2 = 16
x^2 + y^2 = 16 = (5 - x - y)^2

Expanding the equation on the right side, we get:

x^2 + y^2 = 16 = 25 + x^2 + y^2 - 10x - 10y + 2xy

Rearranging and simplifying, we have:

10x + 10y - 2xy = 9

Now, we have a system of two equations:

x + y + z = 5
10x + 10y - 2xy = 9

To solve this system, we can use different methods such as substitution or elimination. However, in this case, we can observe that the second equation can be rewritten as:

5x + 5y - xy = 4.5

Now, let's solve this equation for one variable and substitute it back into one of the original equations. Let's solve for x:

x = (4.5 - 5y) / (5 - y)

Now, substitute this value of x into the equation x + y + z = 5:

(4.5 - 5y) / (5 - y) + y + z = 5

Simplifying, we get:

4.5 - 5y + y(5 - y) + z(5 - y) = 5(5 - y)

Expand and rearrange the equation to form:

-4y^2 + 10y + yz - 5z = 16

This equation represents the intersection of the two given equations x^2 + y^2 = 16 and x + y + z = 5.

Now that we have the equation for the intersection, we can proceed to find the curvature at the point (0,4,1).

Answer 2

The parametric equations for the circle are