GRAPHS

• Graph 1 — A paraboloid \(f(x,y) = x^2 + y^2\).
• Graph 2 — A cone \(g(x,y) = \sqrt{x^2 + y^2}\).

Background & Notes — Exercise 1

Let \(f(x,y)\) be a function with continuous partial derivatives at a point \(P(a,b)\).

Then \(f(x,y)\) is differentiable at \(P(a,b)\), and the equation of its tangent plane at \(P\) is:

\[z = f_x(a,b)\big(x - a\big) + f_y(a,b)\big(y - b\big) + f(a,b)\]

EXERCISE 1 — Tangent Planes & Differentiability

Assumptions

Let \(f(x,y)\) and \(g(x,y)\) be the functions:

\[\begin{align*}
f(x,y) &= x^2 + y^2\\
g(x,y) &= \sqrt{x^2 + y^2}
\end{align*}\]

And let \(P(a,b)\) be the point \(P(-1,2)\).

The graph of \(f(x,y)\) is a paraboloid.

The graph of \(g(x,y)\) is a cone.

Instructions

1. Compute & Graph — Tangent planes of a paraboloid.

Compute the partial derivatives of \(f(x,y) = x^2 + y^2\).

Using the partial derivatives, identify any points \((x,y)\) at which the paraboloid \(f(x,y)\) is not differentiable and lacks a tangent plane.

Then, construct the equation of the tangent plane to the paraboloid \(f(x,y)\) at the point \(P(-1,2)\).

Go to graph 1 and add the tangent plane at \(P(-1,2)\) to the graph of the paraboloid \(f(x,y) = x^2 + y^2\).

2. Compute & Graph — Tangent planes of a cone.

Compute the partial derivatives of \(g(x,y) = \sqrt{x^2 + y^2}\).

Using the partial derivatives, identify any points \((x,y)\) at which the cone \(g(x,y)\) is not differentiable and lacks a tangent plane.

Then, construct the equation of the tangent plane to the cone \(g(x,y)\) at the point \(P(-1,2)\).

Go to graph 2 and add the tangent plane at \(P(-1,2)\) to the graph of the cone \(g(x,y) = \sqrt{x^2 + y^2}\).

3. Compare

In step 1, using continuity of the partial derivatives, you found that the paraboloid \(f(x,y) = x^2 + y^2\) is differentiable everywhere, and has a tangent plane at all points.

In step 2, you identified the origin \(O(0,0)\) as the only point at which the cone \(g(x,y) = \sqrt{x^2 + y^2}\) is not differentiable and does not have a tangent plane.

Compare the behavior of the graphs of the paraboloid \(f(x,y)\) and the cone \(g(x,y)\) at the origin \(O(0,0)\):

• Describe the difference you see in the two graphs at the origin.
• Can you think of an example of a single-variable function \(y = h(x)\) from single-variable calculus whose graph exhibits behavior similar to the cone?

Big Ideas — Tangent Planes & Differentiability

Differentiability of a function \(f(x,y)\) is equivalent to the existence of a well-defined tangent plane at each point. These tangent planes vary smoothly over the graph of the function.

Differentiability/existence of tangent planes depends only on continuity of the partial derivatives.

Points where a function fails to be differentiable correspond to points where the graph has a cusp, fold, sharp edge, etc.

Background & Notes — Exercise 2

The (local) linearization \(L(x,y)\) of a differentiable function \(f(x,y)\) at a point \(P(a,b)\) is the linear function that best approximates the function’s values at points \((x,y)\) near \(P(a,b)\).

The point \(P(a,b)\) is the center of the approximation.

“Local” refers to the fact that these approximations usually decrease in accuracy as the distance from the center increases.

The graph of a linearization \(L(x,y)\) is a tangent plane, so the equation defining \(L(x,y)\) is the same as the tangent plane:

\[f(x,y) \approx L(x,y) = f_x(a,b)\big(x - a\big) + f_y(a,b)\big(y - b\big) + f(a,b)\]

EXERCISE 2 — Linear Approximation

Assumptions

Consider the value:

\[(2.01)^3(5.02)^2\]

Instructions

1. Model the Expression

Identify a function \(f(x,y)\) that models the expression \((2.01)^3(5.02)^2\).

Then, identify the point \(P(a,b)\) that will serve as a good center for the linearization of \(f(x,y)\).

2. Construct the Linearization

Use the equation:

\[f(x,y) \approx L(x,y) = f_x(a,b)\big(x - a\big) + f_y(a,b)\big(y - b\big) + f(a,b)\]

to construct the linear function \(L(x,y)\) that approximates \(f(x,y)\) at the center \(P(a,b)\).

3. Evaluate & Compare

Use the linear function \(L(x,y)\) to approximate the value \((2.01)^3(5.02)^2\).

Then, compute the value \((2.01)^3(5.02)^2\) directly — either by hand, or using a calculator or other software.

How "good" is the approximation \(L(2.01,5.02)\) compared to the actual value \((2.01)^3(5.02)^2\)?

Big Ideas — Linear Approximation

A differentiable functions can be approximated using a linear function whose graph is a tangent plane.

Accuracy of linear approximations generally decreases as the distance from the center of the approximation increases.