Problem 47 - Calculus Early Transcendentals

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. $$f(x, y)=\sqrt{3} / 2 ; P(1 / 2, \sqrt{3})$$

Step-by-Step Solution

Answer

Answer: The slope of the tangent line to the level curve is \(-\frac{\sqrt{3}}{2}\), and it verifies orthogonality with the gradient vector at point \(P\).

Step 1: Find the gradient (partial derivatives) of the function

We are given the function \(f(x,y) = \sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}\). To find the gradient, we should compute the partial derivatives of the function with respect to both \(x\) and \(y\): $$ \frac{\partial f}{\partial x} = -\frac{x}{4\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}} $$ $$ \frac{\partial f}{\partial y} = -\frac{y}{8\sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}} $$

Step 2: Compute the gradient at the point P

Now that we have the partial derivatives of the function, let's compute the gradient at the given point \(P(1/2, \sqrt{3})\). Insert these coordinate values into both \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\): $$ \frac{\partial f}{\partial x}(1/2, \sqrt{3}) = -\frac{1/2}{4\sqrt{1 - \frac{(1/2)^2}{4} - \frac{(\sqrt{3})^2}{16}}} = -\frac{1}{2\sqrt{1-\frac{1}{8}}} = -\frac{1}{2\sqrt{3}/4} = -\frac{1}{\sqrt{3}/2} = -\sqrt{3} $$ $$ \frac{\partial f}{\partial y}(1/2, \sqrt{3}) = -\frac{\sqrt{3}}{8\sqrt{1 - \frac{(1/2)^2}{4} - \frac{(\sqrt{3})^2}{16}}} = -\frac{\sqrt{3}}{2\sqrt{3}/4} = -2 $$ Thus, the gradient (vector) at point \(P\) is $$\nabla f(1/2, \sqrt{3}) = <-\sqrt{3}, -2>$$.

Step 3: Find the slope of the tangent line at point P

At any point on the level curve, the gradient direction is orthogonal to the tangent line of the level curve. Therefore, if we find the tangent line's slope, it should be the negative reciprocal of the \(y\)-component of the gradient divided by the \(x\)-component (\(-\frac{1}{\frac{\partial f}{\partial y}/\frac{\partial f}{\partial x}}\)): $$ \text{slope of tangent line} = -\frac{1}{\frac{-2}{-\sqrt{3}}} = -\frac{1}{\frac{2}{\sqrt{3}}} = -\frac{\sqrt{3}}{2} $$

Step 4: Verify orthogonality between the gradient and tangent line

To check if the tangent line is orthogonal to the gradient vector, we can compute the dot product between the gradient vector and the tangent vector. Let's find the tangent vector first. Since we have the slope of the tangent line, we can find any other point on the tangent line (to create the tangent vector). Let's consider the points \((1/2 + \Delta x, \sqrt{3} + \Delta y)\), where the slope is the same as that of the tangent line, i.e., \(-\frac{\sqrt{3}}{2}\). $$ \frac{\Delta y}{\Delta x} = -\frac{\sqrt{3}}{2} $$ Let's set \(\Delta x = 1\): $$ \Delta y = -\frac{\sqrt{3}}{2} $$ Thus, the tangent vector is $$<\Delta x, \Delta y> = <1, -\frac{\sqrt{3}}{2}>$$. Now, we compute the dot product between the gradient vector and the tangent vector: $$ \nabla f(1/2, \sqrt{3}) \cdot <1, -\frac{\sqrt{3}}{2}> = <-\sqrt{3}, -2> \cdot <1, -\frac{\sqrt{3}}{2}> = (-\sqrt{3})(1) + (-2)\left(-\frac{\sqrt{3}}{2}\right) $$ $$ = -\sqrt{3} + \sqrt{3} = 0 $$ Since the dot product equals 0, this confirms that the tangent line is orthogonal to the gradient vector at point \(P\).

Key Concepts

Ellipsoid

An ellipsoid is a three-dimensional shape, similar to an elongated sphere. It is defined mathematically as a surface where the equation involves quadratic terms. In this exercise, we have the equation of an ellipsoid given by:
\[ f(x, y) = \sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}}\]This particular equation represents the upper half of an ellipsoid as it includes a square root that yields only non-negative results. In essence, the equation implies a constraint on the values of \(x\) and \(y\) that effectively limits the possible points to those lying on or above the XY-plane.### Key Features of an Ellipsoid:- **Axial Symmetry**: The axes of the ellipsoid need not be the same, which means you can have different lengths along the x, y, and z directions.- **Level Sections**: Cutting an ellipsoid yields an ellipse. This is true even in the exercise where the resultant level curve is more than just a simple line.Understanding the ellipsoid in this context is crucial as it forms the basis from which the problems involving level curves and gradients emerge.

Level Curve

A level curve is a cross-section of a surface that shows points where the function has a constant value. It is particularly useful in discovering the behavior of functions of several variables. In the context of our ellipsoid, the level curve is a contour line on the surface representing points that satisfy the equation:
\[f(x, y) = \frac{\sqrt{3}}{2}\]This means that at any point along this curve, the value of the function \(f(x, y)\) remains the same.### Understanding Level Curves:- **Tangent Lines**: At any given point on a level curve, the tangent is a line that touches the curve at only that point without crossing it.- **Gradient and Tangent Relationships**: The gradient of the function at this point is always perpendicular (orthogonal) to the tangent line of the level curve.Level curves help visualize functions of two variables, making it easier to apprehend the nature of the surfaces they describe.

Partial Derivative

Partial derivatives are foundational in understanding how a multi-variable function changes. They represent the rate of change of the function with respect to one variable while keeping the others constant. In simpler terms, in a function involving several variables, a partial derivative gives you the slope of the function along one dimension while ignoring others' variances.
In this problem, we computed the partial derivatives- **\(\frac{\partial f}{\partial x}\):** Measures the rate at which \(f\) changes as \(x\) changes, holding \(y\) constant.- **\(\frac{\partial f}{\partial y}\):** Measures the rate at which \(f\) changes as \(y\) changes, holding \(x\) constant.### Importance of Partial Derivatives:- **Gradient Formation**: The partial derivatives collectively form the gradient vector of the function, which is always orthogonal to the level curve.- **Optimization**: They are crucial in finding the maxima or minima of functions, frequently used in physics and economics to identify optimal conditions.Understanding partial derivatives gives a deep insight into the multidimensional changes which are often not intuitive on a three-dimensional plane.