Question

If \[F\left( u \right) = f\left( {x,y,z} \right)\] be a homogeneous function of degree \[n\] in \[x,y,z\], then find the value of \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} + z\dfrac{{\partial u}}{{\partial z}}\].
A. \[\dfrac{{F\left( u \right)}}{{F\left( u \right)}}\]
B. \[nF\left( u \right)\]
C. \[nu\]
D. None of these

Answer 1

Hint: First we assume \[p = xt\], \[q = yt\] and \[r = zt\] in \[f\left( {x,y,z} \right)\] and apply the homogeneous function condition. Then find the partial derivative \[f\left( {p,q,r} \right)\] with respect to \[t\]. Again, find the partial differentiation of \[p = xt\], \[q = yt\] and \[r = zt\] with respect to \[t\] and calculate the desired result.

Formula used:
A homogeneous of \[n\] degree of \[x,y,z\] can be written as \[f\left( {tx,ty,tz} \right) = {t^n}f\left( {x,y,z} \right)\].

Complete step by step solution:
Assume that,
\[p = xt\]
\[q = yt\]
 \[r = zt\]
Since \[f\left( {x,y,z} \right)\] is a homogeneous function of degree \[n\] in \[x,y,z\].
So, \[f\left( {tx,ty,tz} \right) = {t^n}f\left( {x,y,z} \right)\]
Substitute \[p = xt\], \[q = yt\] and \[r = zt\]
\[f\left( {p,q,r} \right) = {t^n}f\left( {x,y,z} \right)\]
Given that \[F\left( u \right) = f\left( {x,y,z} \right)\].
So, \[f\left( {p,q,r} \right) = {t^n}F\left( u \right)\]
Partial differentiate with respect to \[t\].
\[\dfrac{\partial }{{\partial t}}f\left( {p,q,r} \right) = \dfrac{\partial }{{\partial t}}\left[ {{t^n}F\left( u \right)} \right]\]
Apply chin rule to left side:
\[ \Rightarrow \dfrac{{\partial f}}{{\partial p}}\dfrac{{\partial p}}{{\partial t}} + \dfrac{{\partial f}}{{\partial q}}\dfrac{{\partial q}}{{\partial t}} + \dfrac{{\partial f}}{{\partial r}}\dfrac{{\partial r}}{{\partial t}} = \dfrac{\partial }{{\partial t}}\left[ {{t^n}F\left( u \right)} \right]\]
Apply the power rule formula
\[ \Rightarrow \dfrac{{\partial f}}{{\partial p}}\dfrac{{\partial p}}{{\partial t}} + \dfrac{{\partial f}}{{\partial q}}\dfrac{{\partial q}}{{\partial t}} + \dfrac{{\partial f}}{{\partial r}}\dfrac{{\partial r}}{{\partial t}} = n{t^{n - 1}}F\left( u \right)\] ……..(i)
Calculate the derivative of \[p = xt\] with respect to \[t\]
\[\dfrac{{\partial p}}{{\partial t}} = \dfrac{\partial }{{\partial t}}\left( {xt} \right)\]
\[ \Rightarrow \dfrac{{\partial p}}{{\partial t}} = x\]
Calculate the derivative of \[q = yt\] with respect to \[t\]
\[\dfrac{{\partial q}}{{\partial t}} = \dfrac{\partial }{{\partial t}}\left( {yt} \right)\]
\[ \Rightarrow \dfrac{{\partial q}}{{\partial t}} = y\]
Calculate the derivative of \[r = zt\] with respect to \[t\]
\[\dfrac{{\partial r}}{{\partial t}} = \dfrac{\partial }{{\partial t}}\left( {zt} \right)\]
\[ \Rightarrow \dfrac{{\partial r}}{{\partial t}} = z\]
Now substitute the value of \[\dfrac{{\partial p}}{{\partial t}}\], \[\dfrac{{\partial q}}{{\partial t}}\], and \[\dfrac{{\partial r}}{{\partial t}}\] in equation (i)
\[x\dfrac{{\partial f}}{{\partial p}} + y\dfrac{{\partial f}}{{\partial q}} + z\dfrac{{\partial f}}{{\partial r}} = n{t^{n - 1}}F\left( u \right)\]
Now put the value of \[p,q,r\] in the above equation.
\[ \Rightarrow x\dfrac{{\partial f}}{{\partial \left( {xt} \right)}} + y\dfrac{{\partial f}}{{\partial \left( {yt} \right)}} + z\dfrac{{\partial f}}{{\partial \left( {zt} \right)}} = n{t^{n - 1}}F\left( u \right)\]
Now put \[t = 1\] in the above equation
\[ \Rightarrow x\dfrac{{\partial f}}{{\partial x}} + y\dfrac{{\partial f}}{{\partial y}} + z\dfrac{{\partial f}}{{\partial z}} = nF\left( u \right)\]
Hence option B is the option.

Note:The equation \[x\dfrac{{\partial f}}{{\partial x}} + y\dfrac{{\partial f}}{{\partial y}} + z\dfrac{{\partial f}}{{\partial z}} = nF\left( u \right)\] is known as the Euler theorem. Euler theorem is applicable only to homogeneous equations. Euler theorem for two variables is \[x\dfrac{{\partial f}}{{\partial x}} + y\dfrac{{\partial f}}{{\partial y}} = nF\left( u \right)\]