Problem 12 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=x y+3 x^{2} ;(1,0)\)

Step-by-Step Solution

Answer

Yes, \(f(x, y)=xy+3x^2\) is differentiable at \((1,0)\).

Step 1: Define Differentiability

A function of two variables, \(f(x, y)\), is differentiable at a point \((a, b)\) if it can be approximated by a linear function near that point. This means that the tangent plane exists and the error of approximation decreases faster than the distance from \((a, b)\) to \((x, y)\).

Step 2: Calculate Partial Derivatives

Calculate the partial derivatives of \(f\) with respect to \(x\) and \(y\).\[\frac{\partial f}{\partial x} = y + 6x, \\frac{\partial f}{\partial y} = x.\]

Step 3: Evaluate Partial Derivatives at the Given Point

Substitute the point \((1, 0)\) into the partial derivatives.\[\frac{\partial f}{\partial x}(1, 0) = 0 + 6 \times 1 = 6, \\frac{\partial f}{\partial y}(1, 0) = 1.\]

Step 4: Formulate Linear Approximation

The linear approximation of \(f(x, y)\) at \((1, 0)\) is given by:\[l(x, y) = f(1, 0) + \frac{\partial f}{\partial x}(1, 0)(x - 1) + \frac{\partial f}{\partial y}(1, 0)(y - 0).\]Calculate \(f(1, 0)\):\[f(1, 0) = 1 \times 0 + 3 \times 1^2 = 3.\]Thus, \[l(x, y) = 3 + 6(x - 1) + 1(y) = 6x + y - 3.\]

Step 5: Verify the Error Term

To show differentiability, the term \(R(x, y)\), defined such that\[f(x, y) = l(x, y) + R(x, y),\]should satisfy \[\lim_{(h, k) \to (0, 0)} \frac{R(1+h, k)}{\sqrt{h^2 + k^2}} = 0.\]Calculate \(R\):\[f(1+h, k) = (1+h)k + 3(1+h)^2,\]\[l(1+h, k) = 6(1+h) + k - 3.\]Subtracting gives:\[R(1+h, k) = (1+h)k + 3(1+h)^2 - 6(1+h) - k + 3.\]Simplify:\[R(1+h, k) = 3h^2 + h^2 - h k + 6h.\]\[\frac{R(1+h, k)}{\sqrt{h^2 + k^2}} = \frac{3h^2 - hk + 6h}{\sqrt{h^2 + k^2}}.\]As \((h, k) \to (0, 0)\), this tends to 0, thus showing that \(f\) is differentiable at \((1, 0)\).

Key Concepts

Partial Derivatives

Understanding partial derivatives is key to dealing with functions of multiple variables. When you have a function like \(f(x, y)\), which has two variables, partial derivatives measure how the function changes as we independently adjust one variable while keeping the other constant. To put it simply, they help us explore the "slopes" or rates of change of the function in the direction of each variable. For \(f(x, y) = xy + 3x^2\), we have two partial derivatives:

• With respect to \(x\): \(\frac{\partial f}{\partial x} = y + 6x\)
• With respect to \(y\): \(\frac{\partial f}{\partial y} = x\)

By substituting the values of the variables at the point of interest, such as \((1, 0)\), into these derivatives, we calculate the slopes in the \(x\)-direction and the \(y\)-direction at that point. This provides crucial information that we use to form the tangent plane.

Tangent Plane

The tangent plane is like a 'flat' surface that touches our 3D function at just a point and lies as close as possible to the surface at that point. It's the three-dimensional equivalent of a tangent line for single-variable functions.For our function \(f(x, y) = xy + 3x^2\), the tangent plane equation is derived using the point \((1, 0)\) and the partial derivatives we found earlier. Specifically, it is expressed as:\[l(x, y) = f(1, 0) + \frac{\partial f}{\partial x}(1, 0)(x - 1) + \frac{\partial f}{\partial y}(1, 0)(y - 0)\]Here, \(l(x, y)\) calculates to \(6x + y - 3\). What this means is, the plane captures how our function behaves right around \((1, 0)\). It provides a "linear snapshot" of the function's behavior very close to this point, hence helping us visualize and simplify complex functions.

Linear Approximation

Linear approximation is a technique that uses the concept of the tangent plane to approximate the values of a function near a given point. It converts our surface into a simpler linear form that's easy to calculate and understand.By using the linear equation of the tangent plane, \(l(x, y) = 6x + y - 3\), we can estimate \(f(x, y)\) for points close to \((1, 0)\). This approximation becomes more and more accurate as the points get closer to \((1, 0)\). Thanks to this approximation, complex calculations that involve curved surfaces are simplified into computations involving planes. This is particularly useful when analyzing small changes or perturbations around a point, allowing for more straightforward and intuitive insights into the function's behavior in that localized area.