Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.7

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.7

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.7

Question 1. If each of the following cases, determine whether the following function is homogeneous or not. If it is so, find the degree.
(i) \( f(x, y) = x^2y + 6x^3 + 7 \)
(ii) \( h(x, y) = \frac { 6x^2y^3 - \pi y^5 + 9x^4y }{ 2020x^2+2019y^2} \)
(iii) \( g(x, y, z) = \frac { \sqrt{3x^2+5y^2+z^2} }{ 4x+7y } \)
(iv) \( U(x, y, z) = xy + \sin(\frac {y^2-2z^2 }{ xy }) \)
Answer:
(i) \( f(x, y) = x^2y + 6x^3 + 7 \)
To check for homogeneity, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = (\lambda x)^2(\lambda y) + 6(\lambda x)^3 + 7 \)
\( = \lambda^2 x^2 \lambda y + 6 \lambda^3 x^3 + 7 \)
\( = \lambda^3 x^2 y + 6 \lambda^3 x^3 + 7 \)
We cannot factor out a common \( \lambda^n \) from all terms because of the constant term '7'. Therefore, this function is not homogeneous.
(ii) \( h(x, y) = \frac { 6x^2y^3 - \pi y^5 + 9x^4y }{ 2020x^2+2019y^2} \)
Replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( h(\lambda x, \lambda y) = \frac { 6(\lambda x)^2(\lambda y)^3 - \pi (\lambda y)^5 + 9(\lambda x)^4(\lambda y) }{ 2020(\lambda x)^2+2019(\lambda y)^2} \)
\( = \frac { 6\lambda^2 x^2 \lambda^3 y^3 - \pi \lambda^5 y^5 + 9\lambda^4 x^4 \lambda y }{ 2020\lambda^2 x^2 + 2019\lambda^2 y^2} \)
\( = \frac { \lambda^5 (6x^2y^3 - \pi y^5 + 9x^4y) }{ \lambda^2 (2020x^2 + 2019y^2)} \)
\( = \lambda^{5-2} h(x, y) \)
\( = \lambda^3 h(x, y) \)
Since we can write \( h(\lambda x, \lambda y) = \lambda^3 h(x, y) \), the function is homogeneous with a degree of 3. Each term in the numerator has a degree of 5 (e.g., \(x^2y^3\) has degree \(2+3=5\)), and each term in the denominator has a degree of 2. The difference in degrees gives the overall degree.
(iii) \( g(x, y, z) = \frac { \sqrt{3x^2+5y^2+z^2} }{ 4x+7y } \)
Replace \( x \) with \( \lambda x \), \( y \) with \( \lambda y \), and \( z \) with \( \lambda z \):
\( g(\lambda x, \lambda y, \lambda z) = \frac { \sqrt{3(\lambda x)^2+5(\lambda y)^2+(\lambda z)^2} }{ 4(\lambda x)+7(\lambda y) } \)
\( = \frac { \sqrt{3\lambda^2 x^2+5\lambda^2 y^2+\lambda^2 z^2} }{ \lambda (4x+7y) } \)
\( = \frac { \sqrt{\lambda^2 (3x^2+5y^2+z^2)} }{ \lambda (4x+7y) } \)
\( = \frac { \lambda \sqrt{3x^2+5y^2+z^2} }{ \lambda (4x+7y) } \)
\( = \lambda^{1-1} g(x, y, z) \)
\( = \lambda^0 g(x, y, z) \)
Since we can write \( g(\lambda x, \lambda y, \lambda z) = \lambda^0 g(x, y, z) \), the function is homogeneous with a degree of 0. When the powers of \( \lambda \) in the numerator and denominator cancel out, the degree is 0.
(iv) \( U(x, y, z) = xy + \sin(\frac {y^2-2z^2 }{ xy }) \)
Replace \( x \) with \( \lambda x \), \( y \) with \( \lambda y \), and \( z \) with \( \lambda z \):
\( U(\lambda x, \lambda y, \lambda z) = (\lambda x)(\lambda y) + \sin(\frac {(\lambda y)^2-2(\lambda z)^2 }{ (\lambda x)(\lambda y) }) \)
\( = \lambda^2 xy + \sin(\frac {\lambda^2 y^2-2\lambda^2 z^2 }{ \lambda^2 xy }) \)
\( = \lambda^2 xy + \sin(\frac {\lambda^2 (y^2-2z^2) }{ \lambda^2 xy }) \)
\( = \lambda^2 xy + \sin(\frac {y^2-2z^2 }{ xy }) \)
In this case, the first term \( \lambda^2 xy \) has a factor of \( \lambda^2 \). However, the second term, \( \sin(\frac {y^2-2z^2 }{ xy }) \), does not have \( \lambda \) as a factor outside of the sine function. Since we cannot factor out a common \( \lambda^n \) from the entire expression, this function is not homogeneous.
In simple words: A function is homogeneous if, when you multiply all its variables by a constant like \( \lambda \), you can pull out that \( \lambda \) raised to some power from the entire function. If there is a constant term or part of the function where \( \lambda \) cannot be factored out, it is not homogeneous. The power of \( \lambda \) you factor out is the degree.

🎯 Exam Tip: Remember that for a function to be homogeneous, every term must have the same total degree. Constant terms or terms within non-algebraic functions (like sin, log) often break homogeneity.

Question 2. Prove that \( f(x, y) = x^3 – 2x^2y + 3xy^2 + y^3 \) is homogeneous. What is the degree? Verify Euler's Theorem for f.
Answer:
First, let's prove that \( f(x, y) \) is homogeneous and find its degree.
Substitute \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = (\lambda x)^3 - 2(\lambda x)^2(\lambda y) + 3(\lambda x)(\lambda y)^2 + (\lambda y)^3 \)
\( = \lambda^3 x^3 - 2\lambda^2 x^2 \lambda y + 3\lambda x \lambda^2 y^2 + \lambda^3 y^3 \)
\( = \lambda^3 x^3 - 2\lambda^3 x^2 y + 3\lambda^3 xy^2 + \lambda^3 y^3 \)
\( = \lambda^3 (x^3 - 2x^2 y + 3xy^2 + y^3) \)
\( = \lambda^3 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^3 f(x, y) \), the function \( f(x, y) \) is homogeneous with a degree of 3. Each term in the original function has a total degree of 3, confirming its homogeneity.

Next, let's verify Euler's Theorem, which states that if \( f \) is a homogeneous function of degree \( n \), then \( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f \). In our case, \( n = 3 \).
First, find the partial derivatives of \( f \) with respect to \( x \) and \( y \):
\( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^3 - 2x^2y + 3xy^2 + y^3) \)
\( = 3x^2 - 4xy + 3y^2 \)

\( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^3 - 2x^2y + 3xy^2 + y^3) \)
\( = -2x^2 + 6xy + 3y^2 \)

Now, calculate \( x \frac{\partial f}{\partial x} \) and \( y \frac{\partial f}{\partial y} \):
\( x \frac{\partial f}{\partial x} = x(3x^2 - 4xy + 3y^2) \)
\( = 3x^3 - 4x^2y + 3xy^2 \)

\( y \frac{\partial f}{\partial y} = y(-2x^2 + 6xy + 3y^2) \)
\( = -2x^2y + 6xy^2 + 3y^3 \)

Add these two expressions together:
\( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = (3x^3 - 4x^2y + 3xy^2) + (-2x^2y + 6xy^2 + 3y^3) \)
\( = 3x^3 - 4x^2y - 2x^2y + 3xy^2 + 6xy^2 + 3y^3 \)
\( = 3x^3 - 6x^2y + 9xy^2 + 3y^3 \)

We can factor out 3 from this result:
\( = 3(x^3 - 2x^2y + 3xy^2 + y^3) \)
\( = 3f(x, y) \)
This matches \( n f(x, y) \) with \( n=3 \). Thus, Euler's Theorem is verified for the given function. This theorem is very useful for checking partial derivative calculations quickly.
In simple words: A function is homogeneous if you can factor out \( \lambda \) to some power when you replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \). The power you get is the degree. Euler's theorem says that if a function is homogeneous, then multiplying its partial derivatives by \( x \) and \( y \) (and adding them) gives you back the original function multiplied by its degree.

🎯 Exam Tip: When verifying Euler's theorem, carefully calculate the partial derivatives first, then perform the multiplications by \( x \) and \( y \), and finally add them. Double-check your factoring to get \( nf \).

Question 3. Prove that \( g (x, y) = x \log(y/x) \) is homogeneous. What is the degree? Verify Euler's Theorem for g.
Answer:
First, let's prove that \( g(x, y) \) is homogeneous and find its degree.
Substitute \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( g(\lambda x, \lambda y) = (\lambda x) \log(\frac{\lambda y}{\lambda x}) \)
\( = \lambda x \log(\frac{y}{x}) \)
\( = \lambda^1 g(x, y) \)
Since \( g(\lambda x, \lambda y) = \lambda^1 g(x, y) \), the function \( g(x, y) \) is homogeneous with a degree of 1. The \( \lambda \) terms in the argument of the logarithm cancel out, leaving a single \( \lambda \) outside.

Next, let's verify Euler's Theorem, which states that if \( g \) is a homogeneous function of degree \( n \), then \( x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} = n g \). In our case, \( n = 1 \).
It's easier to first rewrite \( g(x, y) \):
\( g(x, y) = x (\log y - \log x) \)
\( = x \log y - x \log x \)

Now, find the partial derivatives of \( g \) with respect to \( x \) and \( y \):
\( \frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (x \log y - x \log x) \)
Using the product rule for \( x \log x \): \( \frac{\partial}{\partial x} (x \log x) = (1)\log x + x(\frac{1}{x}) = \log x + 1 \)
So, \( \frac{\partial g}{\partial x} = \log y - (\log x + 1) \)
\( = \log y - \log x - 1 \)

\( \frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (x \log y - x \log x) \)
\( = x (\frac{1}{y}) - 0 \)
\( = \frac{x}{y} \)

Now, calculate \( x \frac{\partial g}{\partial x} \) and \( y \frac{\partial g}{\partial y} \):
\( x \frac{\partial g}{\partial x} = x (\log y - \log x - 1) \)
\( = x \log y - x \log x - x \)

\( y \frac{\partial g}{\partial y} = y (\frac{x}{y}) \)
\( = x \)

Add these two expressions together:
\( x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} = (x \log y - x \log x - x) + x \)
\( = x \log y - x \log x \)
\( = x (\log y - \log x) \)
\( = x \log(\frac{y}{x}) \)
\( = g(x, y) \)
This matches \( n g(x, y) \) with \( n=1 \). Thus, Euler's Theorem is verified for the given function. This shows how Euler's theorem works even with logarithmic functions.
In simple words: A function is homogeneous if, after replacing \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \), you can factor out \( \lambda \) raised to some power. For this function, the degree is 1. Euler's theorem checks if \( x \) times the derivative by \( x \) plus \( y \) times the derivative by \( y \) equals the degree times the original function, which it does.

🎯 Exam Tip: When dealing with logarithmic functions like \( \log(y/x) \), remember the property \( \log(A/B) = \log A - \log B \) to simplify partial differentiation.

Question 4. If \( u(x, y) = \frac { x^2+y^2 }{ \sqrt{x+y} } \), prove that \( x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{3}{2}u \).
Answer:
To prove the given relationship, we will use Euler's Theorem for homogeneous functions. First, we need to determine if \( u(x, y) \) is a homogeneous function and find its degree.
Substitute \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( u(\lambda x, \lambda y) = \frac { (\lambda x)^2+(\lambda y)^2 }{ \sqrt{(\lambda x)+(\lambda y)} } \)
\( = \frac { \lambda^2 x^2+\lambda^2 y^2 }{ \sqrt{\lambda (x+y)} } \)
\( = \frac { \lambda^2 (x^2+y^2) }{ \sqrt{\lambda} \sqrt{x+y} } \)
\( = \lambda^{2 - \frac{1}{2}} \frac { x^2+y^2 }{ \sqrt{x+y} } \)
\( = \lambda^{\frac{3}{2}} u(x, y) \)
Since \( u(\lambda x, \lambda y) = \lambda^{\frac{3}{2}} u(x, y) \), the function \( u(x, y) \) is a homogeneous function of degree \( n = \frac{3}{2} \). The degree is found by subtracting the power of \( \lambda \) in the denominator from the power in the numerator.

According to Euler's Theorem for homogeneous functions, if \( u \) is a homogeneous function of degree \( n \), then:
\( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u \)
Substituting the degree \( n = \frac{3}{2} \) into Euler's Theorem, we get:
\( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{3}{2}u \)
Thus, the given statement is proven. This is a powerful shortcut for derivatives of homogeneous functions.
In simple words: The problem asks us to show a special relationship between a function and its derivatives. We first check if the function is "homogeneous" by replacing \( x \) and \( y \) with \( \lambda x \) and \( \lambda y \). If we can pull out a factor like \( \lambda \) to some power (here, \( \lambda \) to the power of 3/2), then it is homogeneous. Euler's Theorem then directly tells us that \( x \) times the derivative by \( x \) plus \( y \) times the derivative by \( y \) will be equal to this power (3/2) multiplied by the original function.

🎯 Exam Tip: For problems requiring proof of \( x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu \), always start by checking if the function is homogeneous and finding its degree \( n \). Once confirmed, Euler's Theorem directly provides the proof.

Question 5. If \( v(x, y) = \log(\frac{x^2+y^2}{x+y}) \), prove that \( x \frac{\partial v}{\partial x} + y\frac{\partial v}{\partial y} = 1 \).
Answer:
To prove the given relationship, we will use a modified approach involving Euler's Theorem for homogeneous functions. Let's define a new function \( f(x, y) \) such that \( v(x, y) = \log(f(x, y)) \). This means \( f(x, y) = e^{v(x, y)} \).
So, \( f(x, y) = \frac{x^2+y^2}{x+y} \)

Now, let's check if \( f(x, y) \) is a homogeneous function and find its degree.
Substitute \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{(\lambda x)^2+(\lambda y)^2}{(\lambda x)+(\lambda y)} \)
\( = \frac{\lambda^2 x^2+\lambda^2 y^2}{\lambda x+\lambda y} \)
\( = \frac{\lambda^2 (x^2+y^2)}{\lambda (x+y)} \)
\( = \lambda^{2-1} \frac{x^2+y^2}{x+y} \)
\( = \lambda^1 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^1 f(x, y) \), the function \( f(x, y) \) is homogeneous with a degree of \( n = 1 \). This transformation helps us apply Euler's theorem directly.

According to Euler's Theorem for homogeneous functions, if \( f \) is a homogeneous function of degree \( n \), then:
\( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f \)
Substituting \( n = 1 \), we get:
\( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 1 \cdot f \)
\( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = f \)

Now, we need to relate this back to \( v \). We know that \( f = e^v \). Let's find the partial derivatives of \( f \) with respect to \( x \) and \( y \) in terms of \( v \):
\( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (e^v) = e^v \frac{\partial v}{\partial x} \)
\( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (e^v) = e^v \frac{\partial v}{\partial y} \)

Substitute these into the Euler's Theorem equation for \( f \):
\( x (e^v \frac{\partial v}{\partial x}) + y (e^v \frac{\partial v}{\partial y}) = e^v \)
Since \( e^v \) is never zero, we can divide the entire equation by \( e^v \):
\( x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = 1 \)
Thus, the given statement is proven. This shows how Euler's theorem can be adapted for functions involving logarithms by considering their exponential form.
In simple words: We are given a function \( v \) that involves a logarithm. To use Euler's Theorem, we first change the function into an exponential form, calling it \( f \). Then we check if this new function \( f \) is homogeneous and find its degree (which is 1). Euler's Theorem then says that \( x \) times the derivative of \( f \) by \( x \) plus \( y \) times the derivative of \( f \) by \( y \) equals \( f \). By using the chain rule to connect the derivatives of \( f \) and \( v \), we can show that \( x \) times the derivative of \( v \) by \( x \) plus \( y \) times the derivative of \( v \) by \( y \) equals 1.

🎯 Exam Tip: For functions involving logarithms (like \( v = \log f \)), transform them into exponential form (\( f = e^v \)) to apply Euler's theorem. Remember to use the chain rule for derivatives of \( e^v \).

Question 6. If \( w(x, y, z) = \log (\frac{5x^3y^4+7y^2xz^4-75y^3z^4}{x^2+y^2}) \), find \( x \frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} + z\frac{\partial w}{\partial z} \).
Answer:
To find the required expression, we will use Euler's Theorem for homogeneous functions of three variables. First, let's define a new function \( f(x, y, z) \) such that \( w(x, y, z) = \log(f(x, y, z)) \). This means \( f(x, y, z) = e^{w(x, y, z)} \).
So, \( f(x, y, z) = \frac{5x^3y^4+7y^2xz^4-75y^3z^4}{x^2+y^2} \)

Now, let's check if \( f(x, y, z) \) is a homogeneous function and find its degree.
Substitute \( x \) with \( \lambda x \), \( y \) with \( \lambda y \), and \( z \) with \( \lambda z \):
Numerator terms:
1. \( 5(\lambda x)^3(\lambda y)^4 = 5\lambda^3 x^3 \lambda^4 y^4 = 5\lambda^7 x^3 y^4 \)
2. \( 7(\lambda y)^2(\lambda x)(\lambda z)^4 = 7\lambda^2 y^2 \lambda x \lambda^4 z^4 = 7\lambda^7 y^2 x z^4 \)
3. \( -75(\lambda y)^3(\lambda z)^4 = -75\lambda^3 y^3 \lambda^4 z^4 = -75\lambda^7 y^3 z^4 \)
All terms in the numerator have a degree of 7. So, the numerator is homogeneous of degree 7.

Denominator terms:
1. \( (\lambda x)^2 = \lambda^2 x^2 \)
2. \( (\lambda y)^2 = \lambda^2 y^2 \)
All terms in the denominator have a degree of 2. So, the denominator is homogeneous of degree 2.

Therefore, \( f(\lambda x, \lambda y, \lambda z) = \frac{\lambda^7 (5x^3y^4+7y^2xz^4-75y^3z^4)}{\lambda^2 (x^2+y^2)} \)
\( = \lambda^{7-2} \frac{5x^3y^4+7y^2xz^4-75y^3z^4}{x^2+y^2} \)
\( = \lambda^5 f(x, y, z) \)
Since \( f(\lambda x, \lambda y, \lambda z) = \lambda^5 f(x, y, z) \), the function \( f(x, y, z) \) is homogeneous with a degree of \( n = 5 \). This tells us that Euler's theorem will apply here.

According to Euler's Theorem for homogeneous functions of three variables, if \( f \) is a homogeneous function of degree \( n \), then:
\( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = n f \)
Substituting \( n = 5 \), we get:
\( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = 5 f \)

Now, we need to relate this back to \( w \). We know that \( f = e^w \). Let's find the partial derivatives of \( f \) with respect to \( x, y, \) and \( z \) in terms of \( w \):
\( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (e^w) = e^w \frac{\partial w}{\partial x} \)
\( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (e^w) = e^w \frac{\partial w}{\partial y} \)
\( \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (e^w) = e^w \frac{\partial w}{\partial z} \)

Substitute these into the Euler's Theorem equation for \( f \):
\( x (e^w \frac{\partial w}{\partial x}) + y (e^w \frac{\partial w}{\partial y}) + z (e^w \frac{\partial w}{\partial z}) = 5 e^w \)
Since \( e^w \) is never zero, we can divide the entire equation by \( e^w \):
\( x \frac{\partial w}{\partial x} + y \frac{\partial w}{\partial y} + z \frac{\partial w}{\partial z} = 5 \)
This is the required value. Euler's theorem offers a quick way to find this sum without calculating individual partial derivatives explicitly.
In simple words: We have a function \( w \) with a logarithm. To solve this, we first turn it into an exponential function \( f \) so we can use Euler's Theorem. We check the "degree" of this new function \( f \), which is 5. Euler's Theorem tells us that \( x \) times its derivative by \( x \), plus \( y \) times its derivative by \( y \), plus \( z \) times its derivative by \( z \) will be 5 times the function \( f \) itself. By changing the derivatives back to \( w \), we find that the final answer is simply 5.

🎯 Exam Tip: For functions of the form \( w = \log f(x,y,z) \), always convert to \( f = e^w \) and find the degree of \( f \). Euler's theorem then provides a direct path to the solution for \( x \frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} + z\frac{\partial w}{\partial z} \).