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 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. $$f(x, y)=1 / \sqrt{2} ; P(0, \sqrt{8})$$

Short answer

#Problem# We are given the upper half of an ellipsoid function \(f(x, y) = \sqrt{1-\frac{x^2}{4}-\frac{y^2}{16}}\) and a point \(P (0, \sqrt{8})\). Compute the slope of the tangent line to the level curve at point \(P\), and verify that the tangent line is orthogonal to the gradient at this point. #Solution# After computing the partial derivatives, finding the gradient of \(f\) at point \(P\), and computing the slope of the tangent line at point \(P\), we find that the slope of the tangent line at point \(P\) is 0. Also, we verified that the tangent line is indeed orthogonal to the gradient at point \(P\), as their dot product is 0.

Step-by-step solution

Find the derivatives of \(f(x, y)\) with respect to \(x\) and \(y\)

In order to find the tangent line, we first compute the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). $$\frac{\partial f}{\partial x} = \frac{-x}{4\sqrt{1-\frac{x^2}{4}-\frac{y^2}{16}}}$$ $$\frac{\partial f}{\partial y} = \frac{-y}{16\sqrt{1-\frac{x^2}{4}-\frac{y^2}{16}}}$$

Find the gradient of \(f\) at point \(P\)

Next, we evaluate the gradient vector of \(f\) at point \(P (0, \sqrt{8})\). $$\nabla f(P) = \left(\frac{\partial f}{\partial x}(0, \sqrt{8}), \frac{\partial f}{\partial y}(0, \sqrt{8})\right) = \left(0, \frac{-\sqrt{8}}{16\sqrt{1-\frac{0^2}{4}-\frac{8}{16}}} \right) = \left(0, -\frac{1}{\sqrt{2}}\right)$$

Compute the slope of the tangent line to the level curve at point \(P\)

Recall that the level curve is given by \(f(x, y) = \frac{1}{\sqrt{2}}\). Since we are given that point \(P\) lies on the level curve, we can replace \(f(x, y)\) with its equivalent constant value. Now, we can calculate the slope of the tangent line using the relation \(\frac{-\partial f/\partial x}{\partial f/\partial y}\), which is also equal to the negative reciprocal of the partial derivatives at point \(P\). $$slope = \frac{-\frac{\partial f}{\partial x}(P)}{\frac{\partial f}{\partial y}(P)} = -\frac{0}{-\frac{1}{\sqrt{2}}} = 0$$ The slope of the tangent line at point \(P\) is 0.

Verify that the tangent line is orthogonal to the gradient at point \(P\)

To verify that the tangent line is orthogonal to the gradient vector, we need to check if the dot product of the tangent vector and the gradient vector is 0 at point \(P\). The tangent vector can be represented as \((1, slope)\) or \((1, 0)\). Let's compute the dot product of the tangent and gradient vectors: $$dot\_product = \nabla f(P) \cdot (1, 0) = (0, -\frac{1}{\sqrt{2}}) \cdot (1, 0) = 0$$ Since the dot product is 0, we can conclude that the tangent line is indeed orthogonal to the gradient vector at point \(P\).

Key concepts

Partial Derivatives

Partial derivatives are an essential concept in calculus, especially when dealing with functions of multiple variables. They allow us to understand how a function changes as we vary one of its input variables while keeping the others constant. This is similar to taking an ordinary derivative, but in the context of functions with more than one variable.

For a function of two variables, like our ellipsoid function \(f(x, y)\), the partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), measures the rate of change of the function in the direction of the \(x\)-axis. Similarly, \(\frac{\partial f}{\partial y}\) measures the rate of change in the \(y\)-axis direction.

These derivatives are calculated by differentiating the function with respect to one variable, while treating all other variables as constants. In our example, we found:
* \(\frac{\partial f}{\partial x} = \frac{-x}{4\sqrt{1-\frac{x^2}{4}-\frac{y^2}{16}}}\)
* \(\frac{\partial f}{\partial y} = \frac{-y}{16\sqrt{1-\frac{x^2}{4}-\frac{y^2}{16}}}\)

These expressions tell us how rapidly the function \(f(x, y)\) changes as \(x\) and \(y\) vary.

Gradient Vector

The gradient vector is a powerful tool in multivariable calculus that provides information about the direction and rate of the steepest ascent from any given point on a surface. For a function \(f(x, y)\), the gradient vector \(abla f\) is composed of the partial derivatives with respect to \(x\) and \(y\):

\[abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\]

The gradient vector points in the direction of the maximum increase of the function, and its magnitude gives the rate at which the function increases in that direction.

In our specific case, to find \(abla f\) at point \(P(0, \sqrt{8})\), we evaluated the partial derivatives at that point:

\[abla f(P) = (0, -\frac{1}{\sqrt{2}})\]

Thus, this gradient vector indicates that the function decreases most steeply in the direction of the negative \(y\)-axis at point \(P\).

This gradient is crucial for analyzing how the function behaves locally around a point, and it also plays a key role in optimization problems and the study of level curves.

Level Curves

Level curves are an important concept when visualizing and analyzing functions of two variables. A level curve of a function \(f(x, y)\) is the set of points \((x, y)\) where the function takes on a constant value, say \(c\). In mathematical terms, this means \(f(x, y) = c\).

By studying level curves, we can gain insight into the topography of a surface, much like contour lines on a geographical map show elevation.

In our example, the level curve is given by \(f(x, y) = \frac{1}{\sqrt{2}}\), which describes a specific "height" on the ellipsoid.

The significance of level curves becomes especially apparent when looking at tangents and gradients. The tangent line to a level curve at any point is orthogonal (perpendicular) to the gradient vector at that point. This is because the gradient indicates the direction of steepest ascent, whereas the level curve represents a constant value, indicating no change in the function in the direction of the tangent.

In summary, level curves help us understand the geometrical and dynamic properties of surfaces generated by functions of two variables.