Problem 6 - Calculus Early Transcendentals

Question

The level curves of the surface \(z=x^{2}+y^{2}\) are circles in the \(x y\) -plane centered at the origin. Without computing the gradient, what is the direction of the gradient at (1,1) and (-1,-1) (determined up to a scalar multiple)?

Step-by-Step Solution

Answer

Answer: The direction of the gradient at the points (1,1) and (-1,-1) is parallel to the y = x line, and can be expressed as a vector (1,1) up to a scalar multiple.

Step 1: Sketch the level curves

In order to have a better understanding of the gradient's direction at the given points (1,1) and (-1,-1), let's sketch the level curves of the function z=x^2+y^2. The level curves are circles with radii growing with the increase in the function's value. This is because for any constant value C, x^2+y^2=C represents a circle centered at the origin with radius sqrt(C).

Step 2: Determine the level curves passing through the given points

Let's determine the level curves of z=x^2+y^2 passing through the points (1,1) and (-1,-1). For the point (1,1), we have z=(1)^2+(1)^2=2. So, the level curve is a circle with equation x^2+y^2=2 passing through the point (1,1). Similarly, for the point (-1,-1), we have z=(-1)^2+(-1)^2=2. So, the level curve is the same circle with equation x^2+y^2=2 passing through the point (-1,-1).

Step 3: Determine the directions of the gradients

Since the gradient of a scalar function at a point is always perpendicular to the level curve passing through that point, the direction of the gradient at the points (1,1) and (-1,-1) can be determined by finding the directions that are perpendicular to the tangent lines of the level curve at these points. From the sketch of the level curve x^2+y^2=2, it's clear that the tangent lines at the points (1,1) and (-1,-1) are parallel to the y=-x line. Therefore, the direction of the gradient at the points (1,1) and (-1,-1) is parallel to the y=x line. We can express the direction of the gradient as a vector, which is (1,1) up to a scalar multiple.