Problem 46 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given point. $$z=\tan ^{-1}(x+y) ;(0,0,0)$$

Step-by-Step Solution

Answer

The equation of the tangent plane at point (0,0,0) is z = x + y.

Step 1: Find the partial derivatives of the surface equation

To find the equation of the tangent plane, we need to first find the partial derivatives of the given surface equation with respect to x and y. Surface equation: \(z = \tan^{-1}(x + y)\) Compute the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\): \(\frac{\partial z}{\partial x} = \frac{1}{1 + (x + y)^2}\) \(\frac{\partial z}{\partial y} = \frac{1}{1 + (x + y)^2}\)

Step 2: Evaluate the partial derivatives at the given point

Now we need to evaluate these partial derivatives at the given point (0,0,0). Plug in x = 0 and y = 0: \(\left.\frac{\partial z}{\partial x}\right|_{(0,0,0)} = \frac{1}{1 + (0 + 0)^2} = 1\) \(\left.\frac{\partial z}{\partial y}\right|_{(0,0,0)} = \frac{1}{1 + (0 + 0)^2} = 1\)

Step 3: Use the point-slope form of a plane equation to find the tangent plane equation

Now we have all the necessary information to determine the equation of the tangent plane. Using the point-slope form of a plane equation and the partial derivatives evaluated at the given point, we can write the tangent plane equation as follows: \(z - z_0 = \left.\frac{\partial z}{\partial x}\right|_{(0,0,0)} (x - x_0) + \left.\frac{\partial z}{\partial y}\right|_{(0,0,0)} (y - y_0)\) where (x_0, y_0, z_0) is the given point (0,0,0). Plugging in the values, we get: \(z - 0 = 1(x - 0) + 1(y - 0)\) Simplifying the equation, we get:

Step 4: Final Answer

The equation of the tangent plane at point (0,0,0) is: \(z = x + y\)