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

Short answer

Answer: The slope of the tangent line to the level curve at point \(P(\sqrt{2}, 0)\) is \(\frac{1}{2}\).

Step-by-step solution

Compute the gradient \(\nabla f\) at point \(P(\sqrt{2}, 0)\)

To compute the gradient \(\nabla f\), we need to find the partial derivatives of \(f(x,y)\) with respect to \(x\) and \(y\). $$\frac{\partial f}{\partial x} = \frac{-x}{2\sqrt{1-\frac{x^{2}}{4}-\frac{y^{2}}{16}}}$$ $$\frac{\partial f}{\partial y} = \frac{-y}{8\sqrt{1-\frac{x^{2}}{4}-\frac{y^{2}}{16}}}$$ Next, we evaluate these partial derivatives at the point \(P(\sqrt{2}, 0)\): $$\frac{\partial f}{\partial x}(P) = \frac{-(\sqrt{2})}{2\sqrt{1-\frac{(\sqrt{2})^{2}}{4}-\frac{0^{2}}{16}}} = -\frac{1}{2}$$ $$\frac{\partial f}{\partial y}(P) = \frac{-0}{8\sqrt{1-\frac{(\sqrt{2})^{2}}{4}-\frac{0^{2}}{16}}} = 0$$ Thus, the gradient \(\nabla f\) at point \(P\) is \(\left(-\frac{1}{2}, 0\right)\).

Compute the slope of the tangent line

Since the gradient of \(f(x,y)\) at a point is orthogonal to the tangent line to the level curve at the same point, we can find the slope of the tangent line by recognizing that the dot product of the gradient \(\nabla f\) and the direction vector of the tangent line is zero: $$(\nabla f) \cdot (1, m) = 0$$ Using gradient \(\nabla f\) from step 1, we get: $$\left(-\frac{1}{2}, 0\right) \cdot (1, m) = -\frac{1}{2} + 0 \cdot m = 0$$ Solving for \(m\): $$m = \frac{1}{2}$$ So, the slope of the tangent line is \(\frac{1}{2}\).

Verify that the tangent line is orthogonal to the gradient

To verify that the tangent line is orthogonal to the gradient, we need to show that the dot product of the gradient and the direction vector of the tangent line is zero: $$\left(-\frac{1}{2}, 0\right) \cdot (1, \frac{1}{2}) = -\frac{1}{2} + 0 \cdot \frac{1}{2}= -\frac{1}{2} + 0 = 0$$ Since the dot product is indeed zero, the tangent line is orthogonal to the gradient at point \(P(\sqrt{2}, 0)\).

Key concepts

Level Curves

Level curves are a fascinating concept in multivariable calculus. Imagine slicing a 3D surface with horizontal planes; the intersections form these curves. For a function like the ellipsoid given by \( f(x, y) = \sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}} \), level curves are the sets where the function has the same value.
In the context of the exercise, the level curve is a path where the ellipsoid's value remains constant, specifically at \( f(x, y) = \frac{1}{\sqrt{2}} \). This means that the entire curve lies on a plane surface of this particular height.
They help visualize the behavior of multi-variable functions and are essential in understanding how these functions change in different directions.

Partial Derivatives

Partial derivatives are used to differentiate functions of multiple variables with respect to one variable, treating other variables as constants. For the ellipsoid function \( f(x, y) = \sqrt{1 - \frac{x^2}{4} - \frac{y^2}{16}} \), we compute partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) as part of understanding the gradient.
- The partial derivative with respect to \( x \) tells us how the function changes as we adjust \( x \), while keeping \( y \) constant.- Similarly, \( \frac{\partial f}{\partial y} \) shows the function's response to changes in \( y \).
In the given exercise, these calculations are crucial for finding the gradient, which plays a significant role in determining the slope of tangent lines and further analysis concerning the curve's geometry.

Ellipsoid

Ellipsoids in mathematics are surface representations similar to elongated or flattened spheres. They are defined by equations like \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \). In the exercise, the ellipsoid is described in a modified form, representing its upper half.
- The ellipsoid function represents a 3D shape where one-third of the surface is pointed upwards.- The function given restricted to two variables removes the \( z \)-variable's complexity, transforming 3D sections into more manageable 2D curves.
Understanding this framework helps in visualizing the specific curved surface where our calculations apply.

Orthogonality

Orthogonality is a powerful concept meaning "perpendicularity" in geometry. In terms of vectors, two vectors are orthogonal if their dot product is zero. This is particularly useful when talking about tangent lines and gradients.
The gradient vector \( abla f \) is perpendicular to level curves at a given point.
- Using the gradient orthogonal to the level curve ensures that the tangent line to this curve sits directly opposite in angle measures, meaning it's at a right-angle intersection point.
- For instance, calculating the dot product of the gradient \( (-\frac{1}{2}, 0) \) and the tangent's direction vector confirms perfect orthogonality when it equals zero.
This orthogonality confirms the perpendicular relationship which is central to understanding tangent lines in calculus.

Dot Product

The dot product is a method to multiply two vectors, giving a scalar as a result. It combines the magnitudes and the cosine of the angle between them.
In mathematics, if the dot product is zero, it indicates that the vectors are orthogonal (perpendicular).
- In the exercise, the dot product helps verify the orthogonal nature of the gradient to a tangent line.- The formula \( \left(-\frac{1}{2}, 0\right) \cdot (1, m) = 0 \) leads to determining \( m \), the slope of the tangent line.
The dot product not only affirms vectors' relations but is instrumental in proving perpendicularity in geometrical contexts.