Tutorial 1+(pde)

EQT 203 : ENGINEERING MATHEMATICS III Chapter 1 : Partial Differential Equations Tutorial 1 1.

Write down all the first order and second order partial derivatives of the following functions :

( a ) z = 4 x 3 −5 xy 2 + 3 y 3 (b) z = cos( 2 x + 3 y ) ( c ) f = e( x 2 − y 2 ) ( d ) f = x 2 sin(2 x + 3 y ) ∂ 2V ∂ 2V + . ∂x 2 ∂y 2

2.

Let V = ln( x 2 + y 2 ) . Calculate

3.

Let z = sin(3x + 2 y ) . Determine 3

4.

x y Given e  2 x  3 y  e . Find

5.

2 2 If z = xy where x  r  s and y  4 rs , find

6.

If u  x 2  e 3 y and v  2 x  e 3 y , find

7.

Given y 

8.

Determine the stationary points and types of the stationary points for the following surfaces :

∂2 z ∂2z − 2 2 in the simplest form. ∂y 2 ∂x

dy . dx z z and . r s

x x y y , , , u v u v

r2 . Find the percentage of the change in y when r increases by 2%, st 3 s increases by 3% and t increases by 1%.

a)

z  y 2  xy  x 2  4 y  4 x  5

b)

z  y 2  xy  2 x  3 y  6

9.

Given z  x 2  xy  5 y 2  19 y . Find the Hessian of this function.

10.

Solve

u u  3 , given the boundary condition u(0, t )  2e 6t . t x

oooooooooooo

Solution : Question 1 a) ∂z =12 x 2 −5 y 2 ∂x ∂z = −10 xy +9 y 2 ∂y ∂2 z = 24 x ∂x 2 ∂2 z = −10 x +18 y ∂y 2 ∂2 z = −10 y ∂x∂y b)

∂z = −2 sin( 2 x + 3 y ) ∂x ∂z = −3sin( 2 x + 3 y ) ∂y ∂2 z = −4 cos(2 x + 3 y ) ∂x 2 ∂2 z = −9 cos( 2 x + 3 y ) ∂y 2 ∂2 z = −6 cos( 2 x + 3 y ) ∂x∂y

c)

∂f = 2ex ∂x ∂f = −2ey ∂y ∂2 f = 2e ∂x 2 ∂2 f = −2e ∂y 2 ∂2 f =0 ∂x∂y d)

∂f = 2 x[sin(2 x + 3 y ) + x cos(2 x + 3 y )] ∂x ∂f = 3x 2 cos(2 x + 3 y ) ∂y ∂2 f = 2[(1 − 2 x 2 ) sin(2 x + 3 y ) + 4 x cos(2 x + 3 y )] 2 ∂x ∂2 f = −9 x 2 sin(2 x + 3 y ) ∂y 2 ∂2 f = 6 x[cos(2 x + 3 y ) − x sin(2 x + 3 y )] ∂x∂y

Question 2 ∂v 2x = 2 ∂x x + y 2 2y ∂v = 2 ∂y x + y 2 ∂ 2 v 2( y 2 − x 2 ) = ∂x 2 ( x 2 + y 2 ) 2 ∂ 2 v 2( x 2 − y 2 ) = ∂y 2 ( x 2 + y 2 ) 2 Then ∂ 2V ∂ 2V + =0 ∂x 2 ∂y 2

Question 3

∂z = 3 cos(3x + 2 y ) ∂x ∂z = 2 cos(3x + 2 y ) ∂y ∂2z = − 9 sin(3x + 2 y ) ∂x 2 ∂2z = −4 sin(3 x + 2 y ) ∂y 2 Then

3

∂2 z ∂2z − 2 =6z ∂y 2 ∂x 2

Question 4 dy e x y  2   x y dx e 3 Question 5 z  12r 2 s 2  4 s 3 r z  8r 3s  12rs 2 s Question 6 J  6e 3 y ( xe6 y  1) Therefore, x e6 y  u 2  xe6 y  1 x 1  6y v 2( xe  1) y e3 y  u 3( xe6 y  1) y xe3 y  v 3( xe6 y  1)

Question 7 y decreases by 2 %

Question 8 a) (4,-4) is a minimum point. b) (1,-2) is a saddle point. Question 9 H = 19 Question 10 u( x, t )  2e 6t 2 x