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)?

Short answer

Question: Determine the direction of the gradient of the surface \(z = x^2 + y^2\) at the points \((1,1)\) and \((-1,-1)\) without computing the gradient explicitly. Answer: The direction of the gradient at the point \((1,1)\) is parallel to the vector \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\), and the direction of the gradient at the point \((-1,-1)\) is parallel to the vector \(\begin{pmatrix} -1 \\ -1 \end{pmatrix}\).

Step-by-step solution

Visualize the level curve at a point

Draw the level curves of the surface \(z = x^2 + y^2\). For the points \((1,1)\) and \((-1,-1)\), the level curve is a circle with radius \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\). On this circle, both points are at an equal distance from the origin.

Determine the radius of the level curve

Calculate the distance of the point \((1,1)\) from the origin. This is given by the formula \(r = \sqrt{x^2 + y^2}\). In this case, we obtain \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\). Since the level curve is a circle centered at the origin, the point \((-1,-1)\) also lies on a circle with the radius \(\sqrt{2}\).

Find the tangent lines

Find the tangent lines of the level curve at the points \((1,1)\) and \((-1,-1)\). These tangent lines are perpendicular to the radial lines from the origin to the respective points. For the point \((1,1)\), the radial line has a slope of \(\frac{1}{1}=1\). The tangent line will have a slope of \(-1\) since it is perpendicular to the radial line. Similarly, for the point \((-1,-1)\), the radial line has a slope of \(\frac{-1}{-1}=1\), and the tangent line has a slope of \(-1\).

Determine the direction of the gradient

Recall that the gradient of a function at a given point is in the direction of the steepest ascent, which is perpendicular to the level curve at that point. Therefore, the gradient is parallel to the radial lines from the origin to the points \((1,1)\) and \((-1,-1)\). For the point \((1,1)\), the gradient direction will be parallel to the radial line connecting the origin and \((1,1)\), which has a direction vector of \(\begin{pmatrix} 1-0 \\ 1-0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\). For the point \((-1,-1)\), the gradient direction will be parallel to the radial line connecting the origin and \((-1,-1)\), which has a direction vector of \(\begin{pmatrix} -1-0 \\ -1-0 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}\). In conclusion, the direction of the gradient at the point \((1,1)\) is parallel to the vector \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\), and the direction of the gradient at the point \((-1,-1)\) is parallel to the vector \(\begin{pmatrix} -1 \\ -1 \end{pmatrix}\).

Key concepts

Understanding Level Curves

Level curves are an essential concept in understanding functions of two variables. They are essentially the cross-sections of the graph of a function where the function has a constant value. Imagine slicing a three-dimensional surface parallel to the base, each slice represents a level curve in a plane. In the exercise given, the function is \( z = x^2 + y^2 \). The level curves for this function are circles, not just any circles, but those centered at the origin

• For any constant \( c \), the level curve is \( x^2 + y^2 = c \)
• This suggests that for different values of \( c \), you get different circles of different radii
• In the specific context of the points (1,1) and (-1,-1), the level curve is a circle with radius \( \sqrt{2} \)
• This represents all points equidistant from the origin at \( \sqrt{2} \) distance, forming a circle

Directional Derivatives Fundamentals

The directional derivative is incredibly useful, as it provides the rate at which a function changes as you move in a particular direction. It's like asking, "how steep is the hill in this direction?". It's crucial in fields like optimization and computer graphics where moving in the right direction saves resources and time. Here’s how it applies:

• For a function \( f(x, y) \), the directional derivative at a point in the direction of a vector \( \mathbf{v} \) is determined by projecting the gradient of \( f \) onto \( \mathbf{v} \)
• The gradient gives you the steepest ascent, while the directional derivative tells you the rate of ascent in any desired direction
• It's a scalar value reflecting how fast or slow you're climbing or descending

Unveiling Radial Lines

Radial lines are lines that emerge from a single central point and extend outward. Visualize them like spokes on a bicycle wheel. For functions whose level curves are circles centered at the origin, these radial lines are straight-line paths that connect each point to the center. Here's how they fit into our function:

• Radial lines for \( z = x^2 + y^2 \) would move directly from the point back to the origin
• For points like (1,1) or (-1,-1), these lines have direction vectors corresponding exactly to the point coordinates, e.g., \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \)
• Since gradients are perpendicular to level curves, the gradient at any point is often parallel to these radial lines

The Essence of Tangent Lines

Tangent lines provide great insights into the behavior of curves. They "touch" the curve but do not cross it. At any point on a level curve, the tangent line denotes the direction that is parallel to the level curve itself. In simple terms:

• A tangent line at a point represents the instantaneous direction of the curve at that point
• For the function \( z = x^2 + y^2 \), if you're at point (1,1), the slope of the tangent line would be calculated as \( -1 \)
• That's because, for circles, the tangent is perpendicular to the radial line, hence why the slopes are negative reciprocals
• These properties of tangent lines help in defining the exact orientation and gradient directions