Question

Consider the upper half of the ellipsoid $$f(x, y)=\sqrt{1-\frac{x^{2}}{4}-\frac{y^{2}}{16}}$$ and the point P on the given level curve and the point \(P\) on the given level curve of \(f .\) Compute the slope of the line tangent to the level curve at \(P\), and verify that the tangent line is orthogonal to the gradient at that point. $$f(x, y)=\frac{1}{\sqrt{2}} ; P(1,2)$$

Short answer

Based on the solution above, we found that the slope of the tangent line at P(1, 2) is 2. However, the tangent line is not orthogonal to the gradient at point P, as their dot product is not 0.

Step-by-step solution

Compute the gradient of the function f(x, y)

To find the gradient of the function f(x, y), we need to find its partial derivatives with respect to x and y. Let's compute these: $$\frac{\partial f}{\partial x} = \frac{-x}{2\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}}$$ $$\frac{\partial f}{\partial y} = \frac{-y}{8\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}}$$ Now, we have the gradient as: $$\nabla f(x, y) = \left(\frac{-x}{2\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}}, \frac{-y}{8\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}}\right)$$

Evaluate the gradient at point P

Next, we evaluate the gradient at the point P(1, 2) which gives us: $$\nabla f(1,2) = \left(\frac{-1}{2\sqrt{1 - \frac{1}{4} - \frac{4}{16}}}, \frac{-2}{8\sqrt{1 - \frac{1}{4} - \frac{4}{16}}}\right) = \left(\frac{-1}{\sqrt{2}}, \frac{-1}{2\sqrt{2}}\right)$$

Find the slope of the tangent line

To find the slope of the tangent line, we can use the fact that its slope is the negative reciprocal of the gradient in the y-direction divided by the gradient in the x-direction: $$slope = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$ At point P, this yields: $$slope = -\frac{\frac{-1}{\sqrt{2}}}{\frac{-1}{2\sqrt{2}}} = 2$$ Thus, the slope of the tangent line at point P is 2.

Verify if the tangent line is orthogonal to the gradient at P

To check if the tangent line is orthogonal to the gradient at P, we want to see if their dot product is 0. The direction vector of the tangent line is [1, slope] = [1, 2] and the gradient vector is $$\nabla f(1,2) = \left(\frac{-1}{\sqrt{2}}, \frac{-1}{2\sqrt{2}}\right)$$ Now let's find their dot product: $$dot\_product = \left(1, 2\right)\cdot\left(\frac{-1}{\sqrt{2}}, \frac{-1}{2\sqrt{2}}\right) = -\frac{1}{\sqrt{2}} - \frac{2}{2\sqrt{2}} = -\frac{3}{\sqrt{2}}$$ Since the dot product is not 0, the tangent line and gradient are not orthogonal at point P.

Key concepts

Gradient

The gradient is a crucial concept in multivariable calculus that extends the idea of derivatives to functions of several variables. For a function of two variables, like the one in this problem, the gradient is a vector that consists of the partial derivatives of the function with respect to each variable.
The gradient vector points in the direction of the greatest rate of increase of the function. For example, for a function \(f(x, y)\), the gradient is denoted as \(abla f(x, y)\), and it is computed as:

• \(\frac{\partial f}{\partial x}\): the partial derivative of \(f\) with respect to \(x\)
• \(\frac{\partial f}{\partial y}\): the partial derivative of \(f\) with respect to \(y\)

The gradient provides valuable information about the geometry of the function, such as the direction of steepest ascent and helps in determining the nature of level curves.

Partial Derivatives

Partial derivatives highlight how a function changes as each variable changes, while other variables remain constant. In practice, they are the cornerstone of finding the slope in multivariable functions.
In this exercise, to find the partial derivatives of the function \(f(x, y)\), we treat one variable as constant and differentiate with respect to the other. The partial derivatives we found are:

• \(\frac{\partial f}{\partial x} = \frac{-x}{2\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}}\)
• \(\frac{\partial f}{\partial y} = \frac{-y}{8\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}}\)

By evaluating these derivatives at a specific point, we learn how changes in each variable separately affect the output of the function at that point. This information is fundamental when calculating the gradient.

Tangent Line

The tangent line to a curve at a given point provides the best linear approximation to the curve near that point. It is defined by its slope and it gives us an idea of the curve's direction at that exact location.
Finding the slope of the tangent line involves using the partial derivatives we calculated, specifically their relationship as explained in the problem. The slope of the tangent line at a point \((x_0, y_0)\) on a level curve is given by the formula:

• \(slope = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}\)

For the point \(P(1, 2)\), this approach yields a slope of \(2\). Understanding the concept of the tangent line is essential for grasping how functions behave locally around specific points.

Level Curve

A level curve of a function with two variables is a curve in the \(x-y\) plane along which the function has a constant value. Exploring level curves allows us to represent multivariable functions in two dimensions.
In this case, the level curve of the function \(f(x, y)\) for which \(f(x, y) = \frac{1}{\sqrt{2}}\) means we are looking for the set of points \((x, y)\) where the function's value remains constant.

• Level curves are useful in visualizing functions of several variables because they highlight how values change and remain steady across a plane.
• They also provide insights into the topographical features of the function, helping in pair with the gradient to predict behavior around critical points.

Understanding the interplay between level curves and the gradient is crucial in connecting geometric interpretations with analytic results.