Problem 8 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 12 , find an equation of the tangent plane and equations of the normal line to the given surface at the indicated point. \(z=x^{1 / 2}+y^{1 / 2} ;(1,1,2)\)

Step-by-Step Solution

Answer

The equation of the tangent plane is \(z = \frac{1}{2}x + \frac{1}{2}y + 1\) and the equations of the normal line are \(x = 1 + \frac{t}{2}\), \(y = 1 + \frac{t}{2}\), \(z = 2 - t\).

Step 1: Find the Partial Derivatives

First, find the partial derivatives of the surface equation with respect to both x and y. This is required to get the gradient, which will help in finding the tangent plane.For the given surface equation: \(z = x^{1/2} + y^{1/2}\), the partial derivatives are:\(\frac{\partial z}{\partial x} = \frac{1}{2}x^{-1/2}\), and \(\frac{\partial z}{\partial y} = \frac{1}{2}y^{-1/2}\)

Step 2: Evaluate the Partial Derivatives at the Point (1,1,2)

Evaluate the partial derivatives found in Step 1 at the given point (1,1,2).\(\left. \frac{\partial z}{\partial x} \right|_{(1,1)} = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}\), and \(\left. \frac{\partial z}{\partial y} \right|_{(1,1)} = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}\)

Step 3: Write the Equation of the Tangent Plane

Use the point (1,1,2) and the partial derivatives to write the tangent plane equation:\(z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\)Plugging in the values, we get:\(z - 2 = \frac{1}{2}(x - 1) + \frac{1}{2}(y - 1)\)Simplify to get the equation of the tangent plane:\(z = \frac{1}{2}x + \frac{1}{2}y + 1\)

Step 4: Find the Gradient Vector

The gradient vector at point (1,1,2) is given by:\(abla f = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \right)\)Evaluating this at (1,1), we get:\(abla f = \left( \frac{1}{2}, \frac{1}{2}, -1 \right)\)

Step 5: Write the Equations of the Normal Line

Using the point (1,1,2) and the direction vector from the gradient, write the parametric equations for the normal line:\(x = x_0 + t\left( \frac{1}{2} \right)\)\(y = y_0 + t\left( \frac{1}{2} \right)\)\(z = z_0 + t(-1)\)The equations then become:\(x = 1 + \frac{t}{2}\)\(y = 1 + \frac{t}{2}\)\(z = 2 - t\)

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, like our surface equation: \(z = x^{1/2} + y^{1/2}\), partial derivatives help us understand how the function changes as we tweak each variable. If we hold one variable constant and differentiate with respect to the other, we get the partial derivative. For our equation, the partial derivative with respect to \(x\) is \( \frac{\partial z}{\partial x} = \frac{1}{2} x^{-1/2} \) and with respect to \(y\), it's \( \frac{\partial z}{\partial y} = \frac{1}{2} y^{-1/2} \). These derivatives are crucial as they give us the rate of change of the function along the x and y directions, respectively, and they are the first steps towards finding the tangent plane and the normal line.

Gradient Vector

The gradient vector, \(abla f\), is a multi-variable generalization of the derivative. It's a vector that points in the direction of the greatest rate of increase of the function and is composed of the partial derivatives found earlier. For our function, the gradient vector at any point \((x, y)\) is given by \(abla f = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1 \right)\). At the point (1,1), the partial derivatives are both \(\frac{1}{2}\), hence the gradient vector is: \(abla f = (\frac{1}{2}, \frac{1}{2}, -1)\). This vector is essential as its components are used to define both the tangent plane and the normal line.

Normal Line

The normal line to the surface at a point is a line that is perpendicular, or 'normal', to the tangent plane at that point. To find it, we use the point of tangency and the direction given by the gradient vector. The parametric equations for the normal line at (1,1,2) are derived using the gradient vector components: \(x = 1 + \frac{t}{2}\), \(y = 1 + \frac{t}{2}\), and \(z = 2 - t\). Here, \(t\) is a parameter that varies, providing any point along the line. This normal line equation is critical as it can help visualize the orientation of the surface at the given point.