Hi there I need help with one of the problem from my homework package. This week for my calculus class, we were learning how to do partial derivative and finding the tangent of the plane. So I do not know how to approach this problem. Please help me by how to derive this problem:
Let z=xsin(x/y), Which y cannot equal to zero. Show that the tangent plane at any point of the surface passes through the origin of (0,0,0).
To show that the tangent plane at any point of the surface passes through the origin, you need to find the equation of the tangent plane and then check if the point (0, 0, 0) satisfies that equation.
First, let's find the partial derivatives of the function z = x * sin(x/y) with respect to x and y.
To find ∂z/∂x, we treat y as a constant and differentiate x * sin(x/y) with respect to x. Applying the product rule, we get:
∂z/∂x = sin(x/y) + x * (∂sin(x/y)/∂x)
Now, to find ∂sin(x/y)/∂x, we treat x/y as a single variable, say u, and differentiate sin(u) with respect to u. Using the chain rule, we have:
∂sin(x/y)/∂x = (∂sin(u)/∂u) * (∂u/∂x) = cos(u) * (1/y)
Substituting u = x/y, we have:
∂sin(x/y)/∂x = cos(x/y) * (1/y)
Substituting this back into the expression for ∂z/∂x, we get:
∂z/∂x = sin(x/y) + x * (cos(x/y) * (1/y)) = sin(x/y) + (x * cos(x/y))/y
Now, let's find ∂z/∂y. We treat x as a constant and differentiate x * sin(x/y) with respect to y. Applying the quotient rule, we have:
∂z/∂y = x * [(y * cos(x/y) - x * sin(x/y))/(y^2)]
To find the equation of the tangent plane at any point (x0, y0, z0) on the surface, we use the point-normal form of the equation of a plane. The normal vector to the tangent plane is given by the gradient of z at that point:
∇z = (∂z/∂x, ∂z/∂y, -1)
So, at any point (x0, y0, z0), the equation of the tangent plane is:
∂z/∂x * (x - x0) + ∂z/∂y * (y - y0) + (-1) * (z - z0) = 0
Now, substituting the expressions we found for ∂z/∂x and ∂z/∂y, the equation of the tangent plane becomes:
[sin(x0/y0) + (x0 * cos(x0/y0))/y0] * (x - x0) + [x0 * [(y0 * cos(x0/y0) - x0 * sin(x0/y0))/(y0^2)]] * (y - y0) - (z - z0) = 0
Finally, to show that the tangent plane passes through the origin (0, 0, 0), substitute x = 0, y = 0, and z = 0 into the equation of the plane. If the equation holds true, then we can conclude that the tangent plane passes through the origin.