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.

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