Chapter 2 Linear Algebra

MA2001 Differential Calculus of Multivariable Functions

1 Introduction

Quantities of interest in real problems usually depend on more than one variable.

Examples

(a) The height of the ground at map coordinates (x, y) might be expressed mathematically as Height =h(x, y).

(b) The temperature in a lecture room varies with position and time,Temperature =T(x, y, z, t).

(c) The volume of a rectangular box of length x, width y and heightz is V(x, y, z) =xyz.

(d) The magnitude of the electric potential u(x, y, z) due to a uniformly charged sphere with total charge Q isu(x, y, z) = _4πε ^Q

0(x²+y²+z²), x²+y²+z² > a² , where a is the radius of the sphere.

These are all real-valued functions of real variables. In general we have z =f(x₁, x₂,· · · , x_n)≡f ~x^T

, where x₁, x₂,· · · , x_n are the independent variables and z is the dependent variable. f ~x^T

is a scalar field.

It is possible to visualize a function of two variables (a surface in 3-D space) by drawing its graph.

One way is to draw a contour map. A level curve off(x, y) is a set of points (x, y) for which f(x, y) = c, where cis a constant.

A collection of level curves then forms a contour map (as in the familiar contour maps produced by map makers).

Chapter 2

Level curves of a function are defined in the same manner as one obtains contour lines for a topographical map.

Contour map (b)

Contour map

Same level curves for the function f(x, y) = x²+y²

Level curves shown on the graph off(x, y) =x²+y² (c)

Level curves for the functionf(x, y) =x²−y²

Some level curves on the graph of f(x, y) = x²−y²

Computer generated graphs usually show lines for which x is constant and lines for which y is constant in order to give a picture of the surface.

Examples

Part (a) shows a computer-generated graph of z =x²−y². Part (b) shows this graph with level curves lifted to it.

2 Limit, Continuity and Partial Derivative

The concepts of limit, continuity and derivative naturally generalize from the one variable situation to two or more variables.

A function f(x, y) is said to possess a limit f0 at the point (x0, y0) if for anyε >0 it is possible to find a δ >0 such that whenever 0<

q

(x−x₀)²+ (y−y₀)² < δ, that is, (x, y) is in disc of radius δ centered at (x₀, y₀), we have|f(x, y)−f₀|< ε. We write lim

(x,y)→(x0,y0)f(x, y) = f₀ or lim

x→x0 y→y0

f(x, y) = f₀. Intuitively, the above says that when (x, y) is approaching (x0, y0), f(x, y) is approaching f0 then the function f(x, y) is said to possess a limit at the point (x₀, y₀).

Note that the limit exists only if,f(x, y) is approaching f₀ independently of the manner in which (x, y) is approaching (x0, y0).

Example lim

(x,y)→(1,2)(x²+ 3y²+xy) = 1²+ 3×2²+ 1×2 = 15

Example

Show thatf(x, y) = _x2^x+y^{2 2} has no limit at the origin.

Solution:

As (x, y)→(0,0) along the liney = 0, f(x,0) = ^x_x²2 = 1 and the limiting value is 1.

As (x, y)→(0,0) along the liney = 0, f(0, y) and the limiting value is 0.

Hence lim(x,y)→(0,0) x²

x²+y² does not exist.

z = _x2^x+y^{2 2},−1≤x≤1,−1≤y≤1

Example Show lim

(x,y)→(0,0) x²

(x²+y²)¹² = 0 Solution:

Now, 0≤ ^x2

(x²+y²)¹² ≤ ^x2+y2

(x²+y²)¹² = (x²+y²)¹². We observe that lim

(x,y)→(0,0)(x² +y²)¹² = 0. Therefore, lim

(x,y)→(0,0) x²

(x²+y²)¹² = 0.

A function f(x, y) is said to be continuous at the point (x₀, y₀) if and only if it possesses a limit at (x₀, y₀) and lim

(x,y)→(x0,y0)f(x, y) =f(x₀, y₀).

Examples

(a) f(x, y) =x²+ 3y²+xy is continuous everywhere.

Solution:

(b) f(x, y) = _x2^x+y^{2 2} is continuous everywhere except at the origin where f(x, y) is not defined.

Solution:

If then (x₀, y₀)6= (0,0) is defined at (x₀, y₀) and f(x₀, y₀) = _x2^x20

0+y²₀. In addition,

(x,y)→(xlim₀,y0)f(x, y) = lim

(x,y)→(x₀,y0) x²

x²+y² = _x2^x20

0+y²₀ =f(x₀, y₀). (c) f(x, y) =











x²

(x²+y²)¹² (x, y)6= (0,0) A (x, y) = (0,0)

is continuous everywhere iff A= 0.

Solution:

Since lim

(x,y)→(0,0) x²

(x²+y²)¹² = 0, f(x, y) =











x²

(x²+y²)¹² (x, y)6= (0,0) A (x, y) = (0,0)

is continuous everywhere iff A= 0.

Example

Give an example of a functionf(x, y) such that f(x, y) has the same limit asf(0,0) when (x, y) approaches (0,0) along any straight line passing through the origin, however, the function is not continuous at (0,0).

Solution:

Consider the function:

f(x, y) =











2xy²

x²+y⁴ if (x, y)6= (0,0) 0 if (x, y) = (0,0) .

Suppose (x, y) is approaching (0,0) along the line x=tcosθ, y =tsinθ where θ is an angle made counterclockwise from the positive direction of x-axis to the line x=tcosθ, y =tsinθ and t is any real number.

Then if cosθ 6= 0 lim

(x,y)→(0,0) x=tcosθ y=tsinθ

f(x, y) = lim

t→0f(tcosθ, tsinθ) = lim

t→0

2t³cosθsin²θ

t²cos²θ+t⁴sin⁴θ = lim

t→0

2t3 cosθsin2θ t2 t2 cos2θ+t4 sin4θ

t2

= lim

t→0

2tcosθsin²θ

cos²θ+t²sin⁴θ = _cos⁰2θ = 0 =f(0,0)

If cosθ= 0, then this is the case that (x, y) is approaching (0,0) along y-axis.

It is obvious that lim

(x,y)→(0,0) x=0

f(x, y) = lim

y→0f(0, y) = lim

y→00 = 0 =f(0,0).

However, if is (x, y) approaching (0,0) along the curve x=t², y =t where t is any real number, we have lim

(x,y)→(0,0) x=t²

y=t

f(x, y) = lim

t→0f(t², t) = lim

t→0 2t⁴

t⁴+t⁴ = 16= 0 =f(0,0).

3 Partial Differentiation

Recall that for a function f(x) of one variable

df

dx(x₀) = lim

h→0

f(x0+h)−f(x0)

h or _dx^df (x₀) = lim

x→x₀

f(x)−f(x0) x−x0

Now consider the functionf(x, y) of two variables. If y is held constant, sayy =y₀, then f(x, y₀) becomes a function of one variable, x and we may define its derivative at x0,

the partial derivative of f with respect to x at the point (x₀, y₀) as

∂f

∂x(x₀, y₀) = lim

h→0

f(x0+h,y0)−f(x₀,y0)

h or ^∂f_∂x(x₀, y₀) = lim

x→x₀

f(x,y0)−f(x₀,y0) x−x0

Similarly, the partial derivative of f with respect toy at the point (x0, y0) as

∂f

∂y (x₀, y₀) = lim

k→0

f(x0,y0+k)−f(x0,y0)

k or ^∂f_∂y(x₀, y₀) = lim

y→y₀

f(x0,y)−f(x0,y0) y−y0

Note:

(a) A partial derivative is just an ordinary derivative with respect to the variable which is not held constant.

Figure 1: AB− curve of f(x, y₀),CD− curve of f(x₀, y) and similarly for ^∂f_∂y(x, y) =f_y(x, y) = ^∂z_∂y(x, y) = z_y(x, y).

(c) The definition of a partial derivative only requires the idea of a limit of a function of a single variable.

Example

Find the first partial derivatives of f(x, y) = x²y+y³sinx.

Solution:

f(x, y) = x²y+y³sinx⇒ ^∂f_∂x = 2xy+y³cosx,^∂f_∂y =x²+ 3y²sinx

Example

Evaluate ^∂f_∂x(1,2) for f(x, y) = x²y+y³sinx Solution:

Method 1:

Method 2:

f(x, y) = x²y+y³sinx⇒ ^∂f_∂x = 2xy+y³cosx.Then ^∂f_∂x(1,2) = (2xy+y³cosx)|^x=1

y=2 = 4 + 8 cos 1.

Example

Find the first partial derivatives of f(x, y, z) = e^2zcosxy.

Solution:

f(x, y, z) =e^2zcosxy⇒ ^∂f_∂x =−e^2zysinxy,^∂f_∂y =−e^2zxsinxy,^∂f_∂z = 2e^2zcosxy.

We may differentiate the first partial derivatives of a function again to obtain second partial derivatives.

f_xx = _∂x∂x^∂2f = _∂x^{∂ ∂f}_∂x

, f_yy = _∂y∂y^∂2f = _∂y^∂

∂f

∂y

, f_xy = _∂y∂x^∂2f = _∂y^{∂ ∂f}_∂x

, f_yx = _∂x∂y^∂2f = _∂x^∂

∂f

∂y

. Higher order partial derivatives may be similarly defined.

Example

Letf(x, y) = log (x²+y²)¹², find its partial derivatives up to order two.

Solution:

f(x, y) = log (x²+y²)¹² ⇒f_x = ^∂f_∂x = _x2+y^{x 2}, f_y = ^∂f_∂y = _x2+y^{y 2}

⇒f_xx = ^∂_∂x^2f2 = _∂x^{∂ ∂f}_∂x

= _∂x^∂

x x²+y²

= _x2+y^{1 2} − ^2x2

(x²+y²)² = ^y2−x2

(x²+y²)²

, fyy = ^∂_∂y^2f2 = _∂y^∂ ∂f

∂y

= ^x2−y2

(x²+y²)², fxy = _∂y^{∂ ∂f}_∂x

= _∂y^∂ x

x²+y²

= ^−2xy

(x²+y²)²

f_yx = _∂x^∂

∂f

∂y

= _∂x^∂

y x²+y²

= ^−2xy

(x²+y²)²

Note that in the above example fxy =fyx, which is the case for most functions that are useful in engineering and science. However, it is not always true.

f(x, y) =











xy(^x2−y2)

x²+y² if (x, y)6= (0,0) 0 if (x, y) = (0,0)

(a)If (x, y)6= (0,0), findf_xy(x, y), f_yx(x, y) and verify that they are equal.

(b)Show that f_x(0,0) =f_y(0,0).

(c)Show that f_xy(0,0), f_yx(0,0) exist but they are not equal.

Solution:

(a)

There is a theorem saying that if f_xy(x, y)&f_yx(x, y) exist at (x₀, y₀) and nearby and f_xy(x, y)&f_yx(x, y) both are continuous at (x₀, y₀) thenf_xy(x₀, y₀) = f_yx(x₀, y₀).

Now, if (x, y)6= (0,0) f_xy(x, y) = _∂y^∂

∂

∂x

xy(^x2−y2)

x²+y²

= _∂y^∂ h

4x²y³+x⁴y−y⁵ (x²+y²)²

i

= ^{−y6−9x2y4+9x4y2+x6}

(x²+y²)³

fyx(x, y) = _∂x^∂

h−4y²x³−y⁴x+x⁵ (x²+y²)²

i

= ^{−y6−9x2y4+9x4y2+x6}

(x²+y²)³

Sof_xy(x, y) = f_yx(x, y) (b)

fx(0,0) = lim

x→0

f(x,0)−f(0,0)

x = lim

x→0 0−0

x = 0, fy(0,0) = lim

y→0

f(0,y)−f(0,0)

y = lim

x→0 0−0

y = 0 (c)

Ify 6= 0, f_x(0, y) = lim

x→0

f(x,y)−f(0,y)

x = lim

x→0

xy(^x2−y2)

x2+y2 −0

x = lim

x→0

y(^x2−y2)

x²+y² =−y Thus,f_x(0, y) =











−y if y6= 0 0 if y= 0

=−y,∀y f_xy(0,0) = lim

y→0

fx(0,y)−fx(0,0)

y = lim

y→0

−y−0

y =−1

Similarly,

Ifx6= 0, fy(x,0) = lim

y→0

f(x,y)−f(x,0)

y = lim

y→0

xy(^x2−y2)

x2+y2 −0

y = lim

y→0

x(^x2−y2)

x²+y² =x Thus,f_Y(x,0) = x

Forf_xy(x, y) =











−y⁶−9x²y⁴+9x⁴y²+x⁶

(x²+y²)³ if (x, y)6= (0,0)

−1 if (x, y) = (0,0) ,we observe that along y= 0, lim

(x,y)→(0,0)f_xy(x, y) = lim

x→0f_xy(x,0) = 1.

However, alongx= 0, lim

(x,y)→(0,0)f_xy(x, y) = lim

y→0f_xy(0, y) = −1.

It concludes that f_xy(x, y) is not continuous at (0,0).

4 The Chain Rule

For functions of one variable, if y=f(x) and x=x(t), theny=f[x(t)] and ^dy_dt = _dx^dfdx_dt. Iff =f(u) and u=u(x, y) then ^∂f_∂x = _du^df∂u_∂x and ^∂f_∂y = _du^df∂u_∂y.

Example

Iff =f(u) = logu and u= (x² +y²)¹², evaluate ^∂f_∂x,^∂f_∂y. Solution:

∂f

∂x = _du^df∂u_∂x = _u¹ × ¹₂(x²+y²)⁻

1

2 ×2x= ¹

(x²+y²)¹² × ¹₂(x²+y²)⁻

1

2 ×2x= _x2+y^{x 2}

And ^∂f_∂y = _du^{df ∂u}_∂y = _x2+y^{y 2}

For a function of two variables we have

Theorem

Ifz =f(x, y) and x=x(t), y =y(t) then ^dz_dt = ^∂f_∂x^dx_dt + ^∂f_∂y^dy_dt. Proof:

dz

dt = lim

δt→0

z(t+δt)−z(t)

δt = limδt→0 f[x(t+δt),y(t+δt)]−f[x(t),y(t)]

δt .

Now ast change tot+δt, xand y change byδx and δy respectively, where, δx=x(t+δt)−x(t).

Thus the change in f may be expressed as

f[x(t+δt), y(t+δt)]−f[x(t), y(t)] =f(x+δx, y+δy)−f(x, y) =QC =QR+RC =P B+RC

= [f(x+δx, y)−f(x, y)] + [f(x+δx, y+δy)−f(x+δx, y)]

, and

dz

dt = lim

δt→0

z(t+δt)−z(t)

δt = lim

δt→0

f[x(t+δt),y(t+δt)]−f[x(t),y(t)]

δt

= lim

δt→0

f(x+δx,y)−f(x,y)+f(x+δx,y+δy)−f(x+δx,y) δt

= lim

δt→0

hf(x+δx,y)−f(x,y) δx

δx

δt + f(x+δx,y+δy)−f(x+δx,y) δy

δy δt

i

= lim

δx→0

f(x+δx,y)−f(x,y)

δx lim

δt→0 δx

δt + lim

δy→0

f(x+δx,y+δy)−f(x+δx,y)

δy lim

δt→0 δy δt

= ^∂f_∂x^dx_dt + ^∂f_∂y^dy_dt

Example

Letz =f(x, y) = x²+ 2xy and x= sint and y= cost, find ^dz_dt. Solution:

dz

dt = ^∂f_∂x^dx_dt +^∂f_∂y^dy_dt = (2x+ 2y) cost+ 2x(−sint) = (2 sint+ 2 cost) cost+ 2 sint(−sint)

= sin 2t+ 2 cos 2t

This may be confirmed by direct substitution:

z =f(sint,cost) = sin²t+ 2 sintcost⇒ ^dz_dt = 2 sintcost−2 sin²t+ 2 cos²t= sin 2t+ 2 cos 2t

A special case is when z =f(x, y) andy =y(x), that is, x=x and hence z is a function of xonly.

Then _dx^dz = ^∂f_∂x^dx_dx+ ^∂f_∂y^dy_dx = ^∂f_∂x +^∂f_∂ydx^dy.

Example

z =f(x, y) = tan⁻¹

x y

, y = sinx, find ^dz_dx Solution:

dz

dx = ^∂f_∂x^dx_dx+ ^∂f_∂y^dy_dx = ^∂f_∂x +^∂f_∂y^dy_dx = _x2+y^{y 2} − _x2+y^{x 2} cosx= _x2+sin^sinx2x − _x2^x+sin^cosx2x = ^sin_x2^x−x+sin^cos2x^x

Example (Chain rule involving second derivatives) Letz =f(x, y), x=x(t), y =y(t), evaluate ^d_dt^2z2. Solution:

dz

dt = ^∂f_∂x^dx_dt +^∂f_∂y^dy_dt.

d²z

dt² = _dt^{d dz}_dt

= _dt^d

∂f

∂x dx

dt +^∂f_∂y^dy_dt

= ^∂f_∂x^d_dt²2^x +_dt^{d ∂f}_∂xdx

dt +^∂f_∂y^d_dt²2^y + _dt^d

∂f

∂y

dy dt.

We observe that ^∂f_∂x is depending onx and y directly, depending indirectly on t through x and y respectively. To evaluate _dt^{d ∂f}_∂x

we need the Chain Rule and we have

d dt

∂f

∂x

= _∂x^{∂ ∂f}_∂xdx

dt + _∂y^{∂ ∂f}_∂xdy

dt = ^∂_∂x^2f2

dx

dt +_∂y∂x^{∂2f dy}_dt For the same reason, _dt^d

∂f

∂y

= _∂x^∂

∂f

∂y

dx

dt + _∂y^∂

∂f

∂y

dy

dt = _∂x∂y^{∂2f dx}_dt + ^∂_∂y^2f2

dy dt

Then it follows that

d²z

dt² = ^∂f_∂x^d_dt^2x2 +_dt^{d ∂f}_∂xdx

dt + ^∂f_∂y^d_dt^2y2 +_dt^d

∂f

∂y

dy dt

= ^∂f_∂x^d_dt^2x2 +h

∂

∂x

∂f

∂x

_dx

dt +_∂y^{∂ ∂f}_∂xdy

dt

idx

dt + ^∂f_∂y^d_dt²2^y +h

∂

∂x

∂f

∂y

dx

dt +_∂y^∂

∂f

∂y

dy dt

idy dt

= ^∂f_∂x^d_dt^2x2 + ^∂_∂x^2f2

dx dt

2

+ _∂y∂x^{∂2f dy}_dt^dx_dt + ^∂f_∂y^d_dt²2^y + _∂x∂y^{∂2f dx}_dt^dy_dt + ^∂_∂y^2f2

dy dt

²

Example

z =f(x, y), x=t², y =t. Suppose

∂f

∂x(1,1) = ^∂f_∂y(1,1) = ^∂_∂x^2f2(1,1) = ^∂_∂y^2f2(1,1) = e,_∂x∂y^∂2f (1,1) = _∂y∂x^∂2f (1,1) = 2e. Evaluate ^d_dt²2^z fort = 1 Solution:

x=t² ⇒ ^dx_dt = 2t,^d_dt^2x2 = 2, y =t ⇒ ^dy_dt = 1,^d_dt²2^y = 0. Ift = 1, then x= 1, y = 1.

So

d²z dt²

t=1 = ^∂f_∂x(1,1)^d_dt^2x2

t=1+ ^∂_∂x^2f2(1,1) ^dx_dt2

t=1+ _∂y∂x^∂2f (1,1)^dy_dt t=1

dx dt

t=1

+ ^∂f_∂y(1,1)^d_dt^2y2

t=1

+ _∂x∂y^∂2f (1,1)^dx_dt t=1

dy dt

t=1

+ ^∂_∂y^2f2(1,1) ^dy_dt2 t=1

=e×2 +e×4 + 2e×1×2 +e×0 + 2e×2×1 +e×1 = 15e

Theorem (Change of Variable)

Ifz =f(x, y) and x=x(s, t), y =y(s, t), find ^∂z_∂s,^∂z_∂t. Solution:

∂z

∂s = ^∂f_∂x^∂x_∂s + ^∂f_∂y^∂y_∂s, ^∂z_∂t = ^∂f_∂x^∂x_∂t +^∂f_∂y^∂y_∂t

Example

Ifz =f(x, y) = e^2xy and there is the change of variablesx=rcosθ, y =rsinθ, find ^∂z_∂r,^∂z_∂θ. Solution:

∂z

∂r = ∂f

∂x

∂x

∂r + ∂f

∂y

∂y

∂r = 2ye^2xycosθ+ 2xe^2xysinθ = 2re^{r2sin 2θ}sin 2θ and

∂z

∂θ = ∂f

∂x

∂x

∂θ + ∂f

∂y

∂y

∂θ = 2ye^2xy(−rsinθ) + 2xe^2xyrcosθ

= 2r²e^{r2sin 2θ}cos 2θ

Again these results may be checked by substitution and direct partial differentiation.

Example (Chain rule involving second partial derivatives) Letz =f(x, y), x=x(s, t), y =y(s, t), evaluate ^∂_∂t²2^z. Solution:

∂z

∂t = ^∂f_∂x^∂x_∂t + ^∂f_∂y^∂y_∂t

∂²z

∂t² = _∂t^{∂ ∂z}_∂t

= _∂t^∂

∂f

∂x

∂x

∂t +^∂f_∂y^∂y_∂t

= _∂t^{∂ ∂f}_∂x^∂x_∂t

+ _∂t^∂

∂f

∂y

∂y

∂t

= _∂t^{∂ ∂f}_∂x∂x

∂t +^∂f_∂x∂t^{∂ ∂x}_∂t

+ _∂t^∂

∂f

∂y

∂y

∂t +^∂f_∂y∂t^{∂ ∂y}_∂t

We observe that both ^∂f_∂x,^∂f_∂y are depending on x andy directly, depending indirectly on t through x and y respectively. To evaluate _∂t^{∂ ∂f}_∂x

,_∂t^∂

∂f

∂y

we need the Chain Rule and we have

∂

∂t

∂f

∂x

= _∂x^{∂ ∂f}_∂x∂x

∂t +_∂y^{∂ ∂f}_∂x∂y

∂t = ^∂_∂x^2f2∂x

∂t + _∂y∂x^{∂2f ∂y}_∂t

∂

∂t

∂f

∂y

= _∂x^∂

∂f

∂y

∂x

∂t +_∂y^∂

∂f

∂y

∂y

∂t = _∂x∂y^{∂2f ∂x}_∂t +^∂_∂y^2f2

∂y

∂t

It follows that

∂²z

∂t² = _∂t^{∂ ∂f}_∂x∂x

∂t +^∂f_∂x∂t^{∂ ∂x}_∂t

+_∂t^∂

∂f

∂y

∂y

∂t + ^∂f_∂y∂t^{∂ ∂y}_∂t

=

∂²f

∂x²

∂x

∂t +_∂y∂x^{∂2f ∂y}_∂t

∂x

∂t + ^∂f_∂x^∂_∂t²2^x +

∂²f

∂x∂y

∂x

∂t + ^∂_∂y^2f2

∂y

∂t

∂y

∂t +^∂f_∂y^∂_∂t^2y2

= ^∂_∂x^2f2

∂x

∂t

2

+_∂y∂x^{∂2f ∂y}_∂t^∂x_∂t + ^∂f_∂x^∂_∂t²2^x +_∂x∂y^{∂2f ∂x}_∂t^∂y_∂t + ^∂_∂y^2f2

∂y

∂t

2

+^∂f_∂y^∂_∂t^2y2

= ^∂_∂x^2f2

∂x

∂t

2

+ 2_∂x∂y^{∂2f ∂x}_∂t^∂y_∂t + ^∂_∂y^2f2

∂y

∂t

²

+^∂f_∂x^∂_∂t^2x2 +^∂f_∂y^∂_∂t^2y2

Example

Consider z =f(x, y), suppose x=rcosθ, y =rsinθ. Show that

∂²z

∂r² = cos²θ^∂_∂x^2f2 + 2 sinθcosθ_∂x∂y^∂2f + sin²θ^∂_∂y^2f2

Solution:

∂z

∂r = ^∂f_∂x^∂x_∂r + ^∂f_∂y^∂y_∂r = cosθ^∂f_∂x+ sinθ^∂f_∂y

∂²z

∂r² = _∂r^{∂ ∂z}_∂r

= _∂r^∂

cosθ^∂f_∂x + sinθ^∂f_∂y

= cosθ_∂r^{∂ ∂f}_∂x

+ sinθ_∂r^∂

∂f

∂y

= cosθ

∂²f

∂x²

∂x

∂r +_∂y∂x^{∂2f ∂y}_∂r

+ sinθ

∂²f

∂x∂y

∂x

∂r + ^∂_∂y^2f2

∂y

∂r

= cos²θ_∂x^∂22f(x, y) + 2 sinθcosθ_∂x∂y^∂2 f(x, y) + sin²θ_∂y^∂22f(x, y)

Example

Show that in polar coordinates (r, θ) in R², Laplace’s equation ∇²u= ^∂_∂x^2u2 +^∂_∂y^2u2 = 0 takes the form ^∂_∂r^2u2 +_r¹2

∂²u

∂θ² +¹_r^∂u_∂r = 0 Solution:

∂u

∂r = ^∂u_∂x^∂x_∂r + ^∂u_∂y^∂y_∂r

∂u

∂θ = ^∂u_∂x^∂x_∂θ + ^∂u_∂y^∂y_∂θ

x=rcosθ, y =rsinθ⇒ ^∂x_∂r = cosθ,^∂y_∂r = sinθ,^∂x_∂θ =−rsinθ,^∂y_∂θ =rcosθ Thus,

∂u

∂r = ^∂u_∂xcosθ+ ^∂u_∂ysinθ ^∂u_∂θ = ^∂u_∂x(−rsinθ) + ^∂u_∂yrcosθ Therefore,









∂u

∂r

∂u

∂θ









=









cosθ sinθ

−rsinθ rcosθ

















∂u

∂x

∂u

∂y









⇒









∂u

∂x

∂u

∂y









=









cosθ sinθ

−rsinθ rcosθ









−1







∂u

∂r

∂u

∂θ









⇒









∂u

∂x

∂u

∂y









=









cosθ −^sin_r^θ sinθ ^cos_r^θ

















∂u

∂r

∂u

∂θ









So ^∂u_∂x = cosθ^∂u_∂r −^sin_r^θ∂u_∂θ · · ·(∗), ^∂u_∂y = sinθ^∂u_∂r +^cos_r^θ∂u_∂θ · · ·(∗∗)

From (∗∗), we learn that ^∂(f_∂y⁾ = sinθ^∂(f_∂r⁾+ ^cos_r^θ∂(f_∂θ⁾. Notice that the f in the left side of (∗∗) is a

function of x, y, however, the f in the right side of (∗∗) after x, y being replaced by rcosθ and rsinθ, respectively, is a function of r, θ. Let ^∂u_∂y =f, therefore, we have

In addition, we notice that ^∂u_∂y = sinθ^∂u_∂r +^cos_r^θ∂u_∂θ Therefore,

∂²u

∂y² = _∂y^∂

∂u

∂y

= sinθ_∂r^∂ sinθ^∂u_∂r +^cos_r^θ∂u_∂θ

+ ^cos_r^θ_∂θ^∂ sinθ^∂u_∂r +^cos_r^θ∂u_∂θ

= sinθ

sinθ^∂_∂r^2u2 + ^cos_r^θ_∂r∂θ^∂2u − ^cos_r2^θ∂u

∂θ

+ ^cos_r^θ

cosθ^∂u_∂r + sinθ_∂θ∂r^∂2u − ^sinθ_r ^∂u_∂r +^cos_r^θ_∂θ∂r^∂2u

· · ·(1) Similarly,

∂²u

∂x² = ∂

∂x ∂u

∂x

= cosθ ∂

∂r ∂u

∂x

−sinθ r

∂

∂θ ∂u

∂x

= cosθ ∂

∂r

cosθ∂u

∂r − sinθ r

∂u

∂θ

−sinθ r

∂

∂θ

cosθ∂u

∂r − sinθ r

∂u

∂θ

= cosθ

cosθ∂²u

∂r² +sinθ r²

∂u

∂θ − sinθ r

∂²u

∂r∂θ

− sinθ r

cosθ ∂²u

∂θ∂r −sinθ∂u

∂r − sinθ r

∂²u

∂θ² − cosθ r

∂u

∂θ

· · ·(2)

Finally, (1)+(2) we have ^∂_∂x^2u2 + ^∂_∂y^2u2 = ^∂_∂r^2u2 +_r¹2

∂²u

∂θ² +¹_r^∂u_∂r = 0

5 Implicit Functions

Letf(x, y) be a function of two independent variables and supposef_x(x, y)&f_y(x, y) exist and are continuous at (x₀, y₀) and nearby, in addition,f(x₀, y₀) = 0&f_y(x₀, y₀)6= 0. Then about (x₀, y₀) the equation f(x, y) = 0 determines an mplicit function y=y(x) about x0 with y0 =y(x0) and y=y(x) being differentiable atx₀.

The above is also applicable to system of equations:

The system of equations



















f1(x1,· · · , xn, y1,· · · , ym) = 0 ...

f_m(x₁,· · ·, x_n, y₁,· · · , y_m) = 0

of x₁,· · · , x_n and the partial

derivatives ofy₁,· · · , y_m with respect tox_i,1≤i≤n can be found.

Example

The equation x³−xy²+yz²−z³ = 5 determine z as a function of x, y.Find ^∂z_∂x. Solution:

Differentiating x³−xy²+yz²−z³ = 5 with respect tox gives 3x²−y²+ 2zy_∂x^∂z −3z^2∂z_∂x = 0 and ^∂z_∂x = _2zy−3z^y2−3x22 if 2zy−3z² 6= 0

Example

Find the gradient of the tangent _dx^dy at a point (x, y) of the conic

f(x, y) = ax²+ 2hxy+by²+ 2gx+ 2f y+c= 0

Solution:

y is defined as an implicit function of x by the equation