MA 1505 Tutorial 6: Partial Derivatives and Directional Derivative

In the tutorial, we will learn the partial derivatives for multiple variable functions.

Assume $z=f(x,y)$ is a two variable function, then we use the notations to describe the partial derivatives of $f(x,y).$

$f_{x}=\frac{\partial f}{\partial x}$ denotes the partial derivative of f under the variable x.

$f_{y}=\frac{\partial f}{\partial y}$ denotes the partial derivative of f under the variable y.

Similarly, we can also define the second derivative of $f(x,y).$

$f_{xx}=\frac{\partial^{2} f}{\partial x^{2}}$,

$f_{xy}=f_{yx}=\frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial^{2}f}{\partial y\partial x}$  $\text{ if } f(x,y) \text{ is a } C^{2} \text{ function.}$

$f_{yy}=\frac{\partial^{2} f}{\partial y^{2}}$.

Assume $u=(a,b)$ is a unit vector, i.e. its length is 1. If $f(x,y)$ is $C^{1}$ at the point p, then we can define the directional derivative of $f(x,y)$ at point p as

$f_{x}(p) a + f_{y}(p) b$

Theorem 1. Geometric mean is not larger than Arithmetic mean.

For n positive real numbers $a_{1}, a_{2}, ..., a_{n}$,

$(a_{1}...a_{n})^{\frac{1}{n}} \leq \frac{a_{1}+...+a_{n}}{n}$

“=” if and only if $a_{1}=a_{2}=...=a_{n}.$

Theorem 2. Cauchy’s Inequality.

For 2n real numbers $a_{1},..., a_{n}, b_{1},...,b_{n}$,

$(a_{1}b_{1}+...+a_{n}b_{n})^{2}\leq (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})$

“=” if and only if $\frac{a_{1}}{b_{1}}=...=\frac{a_{n}}{b_{n}}.$

Proof.

Method (i). Construct a non-negative function f(x) with respect to variable x

$f(x)= \sum_{i=1}^{n}(a_{i}x-b_{i})^{2}= (\sum_{i=1}^{n} a_{i}^{2}) x^{2} - 2(\sum_{i=1}^{n} a_{i}b_{i}) x + (\sum_{i=1}^{n} b_{i}^{2}).$

Consider the equation f(x)=0, there are only two possibilities: one is the equation f(x)=0 has only one root, the other one is the function has no real roots. Therefore,

$\Delta=4(\sum_{i=1}^{n} a_{i}b_{i})^{2} - 4 ( \sum_{i=1}^{n} a_{i}^{2}) \cdot ( \sum_{i=1}^{n} b_{i}^{2}) \leq 0.$

Hence, $(a_{1}b_{1}+...+a_{n}b_{n})^{2}\leq (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2}).$

Moreover, if “=”, then f(x)=0 has only one root $x_{0}$, i.e. for all $1\leq i \leq n$, $a_{i}x_{0}-b_{i}=0$. That means

$\frac{a_{1}}{b_{1}} =...=\frac{a_{n}}{b_{n}}.$

By the way, the solution of $ax^{2}+bx+c=0$ is $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ and $\Delta=b^{2}-4ac.$

Method (ii). Since $a^{2}+b^{2}\geq 2 ab$, we know

$ab\leq \frac{1}{2}(\lambda^{2} a^{2}+ b^{2}/\lambda^{2})$ for all $\lambda \neq 0.$

Assume $\lambda^{2}=\sqrt{(\sum_{i=1}^{n}b_{i}^{2})/(\sum_{i=1}^{n} a_{i}^{2})}$, for all $1\leq i \leq n$,

$a_{i}b_{i}\leq \frac{1}{2} (\lambda^{2}a_{i}^{2}+b_{i}^{2}/\lambda^{2})$

Take the summation at the both sides,

$\sum_{i=1}^{n} a_{i}b_{i} \leq \frac{1}{2}( \lambda^{2}\sum_{i=1}^{n}a_{i}^{2} + (\sum_{i=1}^{n} b_{i}^{2})/ \lambda^{2})= \sqrt{(\sum_{i=1}^{n} a_{i}^{2}) \cdot (\sum_{i=1}^{n}b_{i}^{2})}.$

Question 1. Assume $u(x,y)$ is a $C^{2}$ function and $u>0$. $u(x,y)$ satisfies the partial differential equation $u u_{xy}= u_{x}u_{y}.$

Prove

(1) $\frac{\partial \ln u}{\partial y}$ is a function of y.

(2) $\frac{\partial \ln u}{\partial x}$ is a function of x.

(3) The solution of $u(x,y)$ has the form $u(x,y)=f(x) g(y)$ for some function $f(x)$ and $g(y)$.

Proof.

(1) Method (i) Make use of derivative.

First, we know $\frac{\partial \ln u}{\partial y}=\frac{u_{y}}{u}$. Second, take the partial derivative of the function with respect to the variable x. That means,

$\frac{\partial }{\partial x} (\frac{u_{y}}{u})= \frac{u_{xy}u- u_{x}u_{y}}{u^{2}}=0$ from the partial differential equation. Therefore, the function $\frac{u_{y}}{u}$ is independent of the variable x. i.e. the function is a function of variable y.

Method (ii) Make use of integration.

Since $u u_{xy}=u_{x}u_{y}$, $\frac{u_{x}}{u}=\frac{u_{xy}}{u_{y}}$, then we take the integration of x at the both sides,

$\int \frac{u_{x}}{u} dx =\int \frac{u_{xy}}{u_{y}} dx$, the left hand side is $\ln u$, the right hand side is $\ln |u_{y}| + h_{1}(y)$ for some function $h(y).$ That means, $\frac{|u_{y}|}{u}=e^{-h_{1}(y)}$. and $\frac{u_{y}}{u}$ is a function of $y$.

(2) is similar to (1).

(3) From part (1), we know $\frac{\partial \ln u }{\partial y}$ is a function of y. Assume $\frac{\partial \ln u }{\partial y} = h_{2}(y)$. Take the integration of y at the both sides, we have

$\ln u= \int h_{2}(y) dy + h_{3}(x)$ for some function $h_{3}(x) .$ $u = e^{\int h_{2}(y) dy} \cdot e^{h_{3}(x)} = g(y) \cdot f(x)$ for some functions $f(x)$ and $g(y).$

Question 2. Assume $L+K=150,$  $L$ and $K$ are non-negative. Find the maximum value of $f(L,K)=50 L^{0.4} K^{0.6}$.

Solution.

Method (i). Langrange’s Method.

$g(L,K,\lambda)=f(L,K)-\lambda(L+K-150)=50L^{\frac{2}{5}}K^{\frac{3}{5}}-\lambda(L+K-150).$

Take three partial derivatives of g,

$\frac{\partial g}{\partial \lambda} = -(L+K-150)=0$

$\frac{\partial g}{\partial L} = 50 \cdot \frac{2}{5} L^{-\frac{3}{5}}K^{\frac{3}{5}} - \lambda$

$\frac{\partial g}{\partial K}= 50 \cdot \frac{3}{5} L^{\frac{2}{5}} K^{-\frac{2}{5}} - \lambda=0$

Solve these three equations, we get $2K=3L$ and $L+K=150$, therefore the maximum value is taken at $L=60$ and $K=90.$

Method (ii). Change to one variable function.

Since L+K=150, we can define the one variable function

$g(L)=f(L,150-L)=50 L^{\frac{2}{5}}(150-L)^{\frac{3}{5}}.$

The derivative of $g^{'}(L)=50 \cdot (\frac{2}{5} L^{-\frac{3}{5}}(150-L)^{\frac{3}{5}} - L^{\frac{2}{5}}\frac{3}{5}(150-L)^{-\frac{2}{5}}).$

The critical point is $L=60.$ The maximal value of g(L) is taken at $L=60, K=90.$

Method (iii). Mathematical Olympic Method.

Use the fact that the geometric mean is not larger than the arithmetic mean.

$f(L,K)=50 L^{\frac{1}{5}}L^{\frac{1}{5}} K^{\frac{1}{5}} K^{\frac{1}{5}} K^{\frac{1}{5}}$

$= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} (3L)^{\frac{1}{5}} (3L)^{\frac{1}{5}} (2K)^{\frac{1}{5}} (2K)^{\frac{1}{5}} (2K)^{\frac{1}{5}}$

$\leq \frac{50}{3^{\frac{3}{5}} 2^{\frac{2}{5}}} \frac{3L+3L+2K+2K+2K}{5}$

$= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} \frac{6(L+K)}{5}$

$= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} \frac{6\cdot 150}{5}$.

The maximum value is taken at $3L=2K.$ i.e. $L=60, K=90.$

Question 3. Assume $24x+18y+12z=144$ and $x, y, z$ are non-negative variables. $f(x,y,z)=18 x^{2} y z$. Find the maximum value of $f(x,y,z).$

Solution.

Method (i). Langrange’s Method

$g(x,y,z,\lambda)=f(x,y,z)-\lambda(24x+18y+12z-144) = 18 x^{2} y z-\lambda(24x+18y+12z-144)$

Take four partial derivatives of $g,$ the critical point is taken at $x=z=1.5y.$ i.e. the maximum value of f(x,y,z) is taken at $x=z=3, y=2.$

Method (ii) Math Olympic Method

$f(x,y,z)=18 x \cdot x \cdot y \cdot z$

$= \frac{18}{12*12*18*12} (12x) \cdot (12x) \cdot (18y) \cdot (12z)$

$\leq \frac{18}{12*12*18*12} (\frac{12x+12x+18y+12z}{4})^{4}$

$= \frac{18}{12*12*18*12} (\frac{144}{4})^{4}$

The maximum value is taken at $12x=12x=18y=12z$, i.e. $x=z=3, y=2.$

Question 4.  2012 Exam MA1505 Semester 1, Question 3(a)

Assume $f(x,y)$ has continuous partial derivatives of all orders, if

$\nabla f = (xy^{2}+kx^{2}y+x^{3}) \textbf{i} + (x^{3}+x^{2}y+y^{2}) \textbf{j},$

Find the value of the constant $k.$

Solution.

Method (i) Use derivatives.

Since $f$ has continuous partial derivative of all orders, $f_{xy}= f_{yx}.$

Since $f_{x}= xy^{2}+kx^{2}y+x^{3}$ and $f_{y}=x^{3}+x^{2}y+y^{2},$

we have

$\frac{\partial}{\partial y} (xy^{2}+kx^{2}y+x^{3})= \frac{\partial}{\partial x} (x^{3}+x^{2}y+y^{2})$

This implies $2xy+kx^{2}=3x^{2}+2xy.$ i.e. $k=3.$

Method (ii). Use integration.

$f_{x}=xy^{2}+kx^{2}y+x^{3} \Rightarrow f(x,y)=\frac{1}{2}x^{2}y^{2} + \frac{k}{3}x^{3}y + \frac{1}{4}x^{4} + h_{1}(y),$

$f_{y}=x^{3}+x^{2}y+y^{2} \Rightarrow f(x,y)=x^{3}y+\frac{1}{2}x^{2}y^{2}+\frac{1}{3}y^{3}+h_{2}(x).$

Comparing them, we know $k=3,$ $h_{1}(y)=\frac{1}{3}y^{3}+C_{1}$ and $h_{2}(x)=\frac{1}{4}x^{4}+C_{2},$ where $C_{1}$ and $C_{2}$ are constants.

Therefore $k=3.$