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An identity Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Short Answer

Expert verified
Answer: When the condition \(ad - bc = 0\) is met, the partial derivatives \(f_x\) and \(f_y\) are both equal to 0 for all \(x\) and \(y\) in the domain of the function \(f\).

Step by step solution

01

Find partial derivatives of \(f(x, y)\)

To find the partial derivative with respect to \(x\), we treat \(y\) as a constant and differentiate the function \(f(x, y)\) using the quotient rule: \(f_x(x, y) = \frac{(cx + dy)(a) - (ax + by)(c)}{(cx + dy)^2}\) Similarly, to find the partial derivative with respect to \(y\), we treat \(x\) as a constant and differentiate the function \(f(x, y)\) using the quotient rule: \(f_y(x, y) = \frac{(cx + dy)(b) - (ax + by)(d)}{(cx + dy)^2}\)
02

Simplify expressions using the given condition

We are given the condition \(ad - bc = 0\), which can also be written as \(bc = ad\). We will now use this condition to simplify the expressions for \(f_x\) and \(f_y\) that we found in the previous step. \(f_x(x, y) = \frac{(cx + dy)(a) - (ax + by)(c)}{(cx + dy)^2} = \frac{acx + ady - acx - bcy}{(cx + dy)^2} = \frac{ady - bcy}{(cx + dy)^2}\) \(f_y(x, y) = \frac{(cx + dy)(b) - (ax + by)(d)}{(cx + dy)^2} = \frac{bcx + bdy - adx - bdy}{(cx + dy)^2} = \frac{bcx - adx}{(cx + dy)^2}\)
03

Use the given condition to show \(f_x = f_y = 0\)

Using the condition \(bc = ad\), we can further simplify the expressions for \(f_x\) and \(f_y\): \(f_x(x, y) = \frac{ady - bcy}{(cx + dy)^2} = \frac{bcy - bcy}{(cx + dy)^2} = \frac{0}{(cx + dy)^2} = 0\) \(f_y(x, y) = \frac{bcx - adx}{(cx + dy)^2} = \frac{adx - adx}{(cx + dy)^2} = \frac{0}{(cx + dy)^2} = 0\) Since we have shown that \(f_x = 0\) and \(f_y = 0\) for all \(x\) and \(y\) in the domain of \(f\), our proof is complete.
04

Explanation

The given condition \(ad - bc = 0\) suggests that the numerator and denominator of the function \(f(x, y)\) are linearly dependent, essentially representing the same information. This causes the partial derivatives with respect to both \(x\) and \(y\) to be zero, which means the function is not changing in any direction (neither increasing nor decreasing) for any point \((x, y)\) in its domain. This is perfectly in line with our result, as \(f_x = f_y = 0\).

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Key Concepts

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

Quotient Rule
When you see a function that looks like a fraction, such as \( f(x, y) = \frac{ax + by}{cx + dy} \), and you need to find its derivative, you'll often apply something called the Quotient Rule. This is a technique in calculus used specifically for differentiating ratios of two functions.

The Quotient Rule states that for a function \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of some variable (like \( x \)), the derivative is given by:
  • \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)
Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

In the exercise, while finding partial derivatives \( f_x \) and \( f_y \), this rule helps us to differentiate the composite expressions formed. For example, treating \( y \) as constant while deriving with respect to \( x \) helps in applying the rule cleanly. This way, it’s manageable to simplify these complex functions to further analyze them.
Linear Dependence
In the context of the given exercise, linear dependence is a key concept. Two vectors or expressions are linearly dependent if one can be expressed as a scalar multiple of the other. Mathematically, for the vectors \( (a, b) \) and \( (c, d) \), linear dependence means that the equation \( ad - bc = 0 \) holds true, which indicates both vectors lie on the same line in the plane.

Apply this idea to understand better why the condition \( ad - bc = 0 \) makes the partial derivatives of the function \( f(x, y) \) zero. The condition implies that the numerators in our simplified expressions for \( f_x \) and \( f_y \) cancel out. This cancellation indicates that changes in \( x \) or \( y \) do not impact \( f(x, y) \), setting the derivatives to zero.

This subtle cancellation effect due to linear dependence suggests an inherent symmetry in the function's numerator and denominator, further accentuating that there is no significant directional change in the function’s behavior over its domain.
Function Analysis
Function Analysis involves breaking down and examining a function's behavior, often through its partial derivatives. By confirming both \( f_x = 0 \) and \( f_y = 0 \), we deduce that the function \( f(x, y) = \frac{ax + by}{cx + dy} \) is constant with respect to both variables. This implies that variations in \( x \) and \( y \) do not alter the function's output across its intended domain.

The conclusion that these partial derivatives are zero hinges on recognizing the linear dependence, an analytical conclusion drawn from geometric interpretations of the vectors formed by the coefficients \( a, b, c, \) and \( d \).

So, function analysis in this context not only confirms the lack of change in \( f(x, y) \) across its domain but also utilizes the properties of both algebraic manipulation and geometric insight. This dual approach enriches our understanding of the underlying symmetries present in function behaviors.

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Most popular questions from this chapter

(Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0\). Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

(1946 Putnam Exam) Let \(P\) be a plane tangent to the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) at a point in the first octant. Let \(T\) be the tetrahedron in the first octant bounded by \(P\) and the coordinate planes \(x=0, y=0\), and \(z=0 .\) Find the minimum volume of \(T\). (The volume of a tetrahedron is one-third the area of the base times the height.)

Identify and briefly describe the surfaces defined by the following equations. $$y=x^{2} / 6+z^{2} / 16$$

Find the points (if they exist) at which the following planes and curves intersect. $$y=2 x+1 ; \quad \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Match equations a-f with surfaces A-F. a. \(y-z^{2}=0\) b. \(2 x+3 y-z=5\) c. \(4 x^{2}+\frac{y^{2}}{9}+z^{2}=1\) d. \(x^{2}+\frac{y^{2}}{9}-z^{2}=1\) e. \(x^{2}+\frac{y^{2}}{9}=z^{2}\) f. \(y=|x|\)

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