/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 Let $$f(x, y)=\left\\{\begin{arr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let $$f(x, y)=\left\\{\begin{array}{ll}{\frac{x y^{2}}{x^{2}+y^{4}},} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ Show that \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist, but \(f\) is not differentiable at \((0,0) .\) (Hint: Use Theorem 4 and show that \(f\) is not continuous at \((0,0) . )\)

Short Answer

Expert verified
The partial derivatives exist but the function is not continuous at \((0,0)\), so it's not differentiable there.

Step by step solution

01

Calculate Partial Derivative with Respect to x

First, calculate the partial derivative of \(f\) with respect to \(x\). The given function is \(f(x, y) = \frac{x y^{2}}{x^{2} + y^{4}}\) when \((x, y) eq (0, 0)\) and \(f(x, y) = 0\) when \((x, y) = (0, 0)\). Use the quotient rule for differentiation:\[ f_x(x, y) = \frac{(x^2 + y^4)(y^2) - x(y^2)(2x)}{(x^2 + y^4)^2} = \frac{y^2(x^2 + y^4 - 2x^2)}{(x^2 + y^4)^2} \]Evaluate at \((0, 0)\):Since \(y = 0\) in the numerator and the denominator is not zero as it approaches zero, \(f_x(0, 0) = 0\).
02

Calculate Partial Derivative with Respect to y

Next, calculate the partial derivative of \(f\) with respect to \(y\). Again start from:\[ f_y(x, y) = \frac{(x^2 + y^4)(2xy) - x(y^2)(4y^3)}{(x^2 + y^4)^2} = \frac{2xy(x^2 + y^4) - 4xy^5}{(x^2 + y^4)^2} \]Simplify:\[ f_y(x, y) = \frac{2xy(x^2 + y^4 - 2y^4)}{(x^2 + y^4)^2} \]Evaluate at \((0, 0)\):This expression simplifies to \(0\) since when \(x=0\), \(xy = 0\), resulting in \(f_y(0, 0) = 0\).
03

Check Continuity at (0,0)

To verify the differentiability condition, check the continuity of \(f(x, y)\) at \((0, 0)\). For \(f\) to be differentiable, it must be continuous at \((0, 0)\). Consider approaching \((0, 0)\) along different paths:1. Along \(x = 0\): \[ f(0, y) = \frac{0 \cdot y^2}{y^4} = 0 \]2. Along \(y = x\): \[ f(x, x) = \frac{x \cdot x^2}{x^2 + x^4} = \frac{x^3}{x^2(1 + x^2)} = \frac{x}{1 + x^2} \] As \(x \to 0\), \(f(x, x) \to 0\).3. Along the path \(y = x^2\): \[ f(x, x^2) = \frac{x (x^2)^2}{x^2 + (x^2)^4} = \frac{x^5}{x^2 + x^8} = \frac{x^3}{1 + x^6} \] As \(x \to 0\), this tends towards \(0\), but varies significantly in behavior concerning other paths.The paths do not yield consistent limit values as they approach \((0,0)\), indicating that \(f\) is not continuous at \((0, 0)\).
04

Conclude Differentiability

Both partial derivatives \(f_x(0,0)\) and \(f_y(0,0)\) exist and are zero. However, due to the lack of continuity when approaching \((0, 0)\) from different directions resulting in different function behaviors, the function \(f(x, y)\) is not continuous at \((0, 0)\). As \(f\) is not continuous at the point, it is not differentiable at \((0,0)\) based on the definition of differentiability which requires continuity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives help us understand how functions change as we vary one variable while keeping others constant. For the function \(f(x, y)\), the partial derivative with respect to \(x\), denoted \(f_x(x, y)\), shows the rate of change of \(f\) when \(x\) changes, holding \(y\) constant.
In the original exercise, we find that \(f_x(0,0) = 0\), which indicates the slope of the function along the \(x\)-axis at the origin is flat. Similarly, \(f_y(0,0) = 0\) indicates the slope along the \(y\)-axis is also flat at this point.
The existence of these partial derivatives shows that the function behaves in a regular, predictable manner in the immediate neighborhood along these specific axes. However, the existence of partial derivatives alone does not guarantee full differentiability of a function at a point.
Continuity
Continuity of a function at a point means you can approach the point from any path and get the same function value. For \(f(x, y)\) to be continuous at \((0,0)\), \(\lim_{(x, y) \to (0,0)} f(x, y)\) must be zero, the same as \(f(0,0)\).
In the exercise, several paths of approach to \((0,0)\) were examined. Along the paths \(x = 0\), \(y = x\), and \(y = x^2\), we see seemingly consistent limits tending toward zero at first. But continuity fails if there’s any inconsistency from different paths, leading to varying limits, making the function not continuous at this point.
Thus, while individual paths may yield limits that suggest continuity, the varying behavior along differing paths implies \(f(x, y)\) is not continuous at \((0,0)\), key for determining differentiability.
Differentiability
Differentiability at a point implies the function has a well-defined tangent plane or, in simpler terms, the function behaves so smoothly at the point that it can be approximated by a linear function. Typically, differentiability requires the function to be continuous and have partial derivatives at the point.
While \(f(x, y)\) does indeed have partial derivatives \(f_x(0,0) = 0\) and \(f_y(0,0) = 0\), the non-continuity at \((0,0)\), as shown by the step-by-step solution, breaks the chain of smooth behavior needed for differentiability. Essentially, if different approaches result in different behaviors, fitting a tangent plane smoothly across all paths is impossible.
This highlights why the existence of partial derivatives alone isn't a guarantee; continuity is equally crucial for ensuring differentiability in multivariable calculus.
Paths of Approach
Paths of approach are important to check how a function behaves as you move closer to a point in various ways. Different paths can reveal different limit behaviors, uncovering the function's complexity.
In the given exercise, different paths like \(x = 0\), \(y = x\), and \(y = x^2\) were used. While initially, these paths seem to suggest that the limit values are consistent, their detailed examination shows anomalies in the behavior of the function \(f(x, y)\) as these paths converge towards \((0,0)\).
Analyzing different routes helps one understand whether the function is continuous or has a consistent behavior at specific points. Ultimately, discrepancies in limit values across paths were critical in proving \(f(x, y)\) isn’t differentiable at \((0,0)\). This path-based method illustrates functions’ nuanced behaviours in multivariable calculus.

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Most popular questions from this chapter

In Exercises \(57-60,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$ f(x, y)=1-x+y-3 x^{2} y, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(1,2) $$

The condition \(\nabla f=\lambda \nabla g\) is not sufficient Although \(\nabla f=\lambda \nabla g\) is a necessary condition for the occurrence of an extreme value of \(f(x, y)\) subject to the conditions \(g(x, y)=0\) and \(\nabla g \neq 0,\) it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of \(f(x, y)=x+y\) subject to the constraint that \(x y=16 .\) The method will identify the two points \((4,4)\) and \((-4,-4)\) as candidates for the location of extreme values. Yet the sum \((x+y)\) has no maximum value on the hyperbola \(x y=16\) . The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum \(f(x, y)=x+y\) becomes.

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} $$

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z, w)=x^{2}+y^{2}+z^{2}+w^{2}\) subject to the constraints \(\quad 2 x-y+z-w-1=0 \quad\) and \(\quad x+y-z+\) \(w-1=0.\)

Tangent curves \(A\) smooth curve is tangent to the surface at a point of intersection if its velocity vector is orthogonal to \(\nabla f\) there. Show that the curve $$ \mathbf{r}(t)=\sqrt{t} \mathbf{i}+\sqrt{t} \mathbf{j}+(2 t-1) \mathbf{k} $$ is tangent to the surface \(x^{2}+y^{2}-z=1\) when \(t=1\)

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