Find the points on the cone \( z^{2}=x^{2}+y^{2} \) that are closest to the point \( (2,1,0) \). Please show your answers to at least 4 decimal places.

To find the points on the cone defined by the equation \( z^2 = x^2 + y^2 \) that are closest to the point \( (2, 1, 0) \), we can use the method of Lagrange multipliers or minimize the distance function directly. ### Step 1: Define the distance function The distance \( D \) from a point \( (x, y, z) \) on the cone to the point \( (2, 1, 0) \) is given by: \[ D = \sqrt{(x - 2)^2 + (y - 1)^2 + z^2} \] Since \( z^2 = x^2 + y^2 \), we can substitute \( z \) in terms of \( x \) and \( y \): \[ D = \sqrt{(x - 2)^2 + (y - 1)^2 + (x^2 + y^2)} \] ### Step 2: Minimize the square of the distance To simplify the calculations, we can minimize the square of the distance \( D^2 \): \[ D^2 = (x - 2)^2 + (y - 1)^2 + (x^2 + y^2) \] Expanding this, we have: \[ D^2 = (x^2 - 4x + 4) + (y^2 - 2y + 1) + (x^2 + y^2) \] Combining like terms: \[ D^2 = 2x^2 + 2y^2 - 4x - 2y + 5 \] ### Step 3: Use the constraint We have the constraint from the cone: \[ z^2 = x^2 + y^2 \] ### Step 4: Set up the Lagrange multipliers We want to minimize \( D^2 \) subject to the constraint \( g(x, y) = z^2 - x^2 - y^2 = 0 \). The Lagrange function is: \[ \mathcal{L}(x, y, z, \lambda) = D^2 + \lambda (g(x, y) - 0) \] ### Step 5: Calculate the gradients We need to find the gradients and set them to zero: 1. \( \frac{\partial \mathcal{L}}{\partial x} = 4x - 4 - 2\lambda x = 0 \) 2. \( \frac{\partial \mathcal{L}}{\partial y} = 4y - 2 - 2\lambda y = 0 \) 3. \( \frac{\partial \mathcal{L}}{\partial z} = 0 \) (since \( z \) is not in \( D^2 \)) 4. \( \frac{\partial \mathcal{L}}{\partial \lambda} = z^2 - x^2 - y^2 = 0 \) ### Step 6: Solve the equations From the first two equations, we can express \( \lambda \): 1. From \( 4x - 4 - 2\lambda x = 0 \): \[ 2\lambda x = 4x - 4 \implies \lambda = \frac{4 - 4/x}{2} = 2 - \frac{2}{x} \] 2. From \( 4y - 2 - 2\lambda y = 0 \): \[ 2\lambda y = 4y - 2 \implies \lambda = \frac{4y - 2}{2y} = 2 - \frac{1}{y} \] Setting the two expressions for \( \lambda \) equal gives: \[ 2 - \frac{2}{x} = 2 - \frac{1}{y} \] This simplifies to: \[ \frac{2}{x} = \frac{1}{y} \implies 2y = x \implies x = 2y \] ### Step 7: Substitute back into the constraint Substituting \( x = 2y \) into the constraint \( z^2 = x^2 + y^2 \): \[ z^2 = (2y)^2 + y^2 = 4y^2 + y^2 = 5y^2 \implies z = \sqrt{5}y \] ### Step 8: Substitute into the distance function Now we can express \( D^2 \) in terms of \( y \): \[ D^2 = 2(2y)^2 + 2y^2 - 4(2y) - 2y + 5 \] \[ = 2(4y^2) + 2y^2 - 8y - 2y + 5 = 8y^2 + 2y^2 - 10y + 5 = 10y^2 - 10y + 5 \] ### Step 9: Minimize \( D^2 \) To minimize \( D^2 \), we take the derivative with respect to \( y \) and set it to zero: \[ \frac{d(D^2)}{dy} = 20y - 10 = 0 \implies y = \frac{1}{2} \] ### Step 10: Find \( x \) and \( z \) Now substituting \( y = \frac{1}{2} \): \[ x = 2y = 2 \cdot \frac{1}{2} = 1 \] \[ z = \sqrt{5}y = \sqrt{5} \cdot \frac{1}{2} = \frac{\sqrt{5}}{2} \] ### Final Points Thus, the point on the cone that is closest to \( (2, 1, 0) \) is: \[ \left( 1, \frac{1}{2}, \frac{\sqrt{5}}{2} \right) \] ### Step 11: Calculate the distance To ensure accuracy, we can calculate the distance: \[ D = \sqrt{(1 - 2)^2 + \left(\frac{1}{2} - 1\right)^2 + \left(\frac{\sqrt{5}}{2} - 0\right)^2} \] \[ = \sqrt{1 + \left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2} \] \[ = \sqrt{1 + \frac{1}{4} + \frac{5}{4}} = \sqrt{1 + \frac{6}{4}} = \sqrt{1 + 1.5} = \sqrt{2.5} = \frac{\