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Chapter 12: Problem 46

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

Short Answer

Expert verified
Answer: The equation of the tangent plane is x + y - z = 0.

Step by step solution

01

Compute the partial derivatives of the surface function.

Before we find the gradient vector, we need to compute the partial derivatives of the surface function with respect to x and y. Let the surface function be $$f(x,y) = \arctan(x+y)$$ Then the partial derivatives are $$\frac{\partial f}{\partial x} = \frac{1}{1+(x+y)^2}\cdot 1\quad \text{and}\quad \frac{\partial f}{\partial y} = \frac{1}{1+(x+y)^2}\cdot 1$$
02

Evaluate partial derivatives at the given point

Now, we need to evaluate the partial derivatives at the given point (0,0,0). Using the partial derivative expressions from Step 1, we have $$\left.\frac{\partial f}{\partial x}\right|_{(0,0,0)} = \frac{1}{1+(0+0)^2} = 1$$ and $$\left.\frac{\partial f}{\partial y}\right|_{(0,0,0)} = \frac{1}{1+(0+0)^2} = 1$$
03

Find the gradient vector at the given point

The gradient vector of the surface function at the given point is obtained by putting together the evaluated partial derivatives. So its components are $$\nabla f(0,0,0) = \left\langle 1,1\right\rangle$$
04

Find the tangent plane equation

Now that we have the gradient vector, we can find the equation of the tangent plane. Let the point on the tangent plane be P = (x,y,z). Then, the tangent plane equation can be found using the normal vector (the gradient vector) and the given point (0,0,0), as follows: $$(x-0)\cdot 1 + (y-0)\cdot 1 + (z-0)\cdot (-1) = 0$$
05

Simplify the tangent plane equation

Finally, we can simplify the tangent plane equation we obtained in Step 4 to get the final equation for the tangent plane: $$x + y - z = 0$$

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