EEM1016 Engineering Mathematics I Tutorial of Partial Derivatives

Trims 1 2014/2015 Basic Concepts (Domain, Range, Graph) 1. Let 2 3( , ) xyf x y x e= .

(a). Evaluate f(2,0). (b). Find and sketch the domain of f.

(c). Find the range of f. [(a).4; (b). 2R ; (c).[0 ] , )∞ 2. Let 2 2 2( , , ) ln(25 ).f x y z x y z= − − − (a). Evaluate . (2, 2, 4)f − (b). Find the domain of f. (c). Find the range of f. [(a) 0, (c). ] ( , ln 25−∞ ]

3. Sketch the graph of the function: (a). (level curve = 0, 2, 4, 6,…) 22 4),( yxyxf +=

(b). (level curve = 3, 4, 5, …) 3),( 22 ++= yxyxf

(c). 229),( yxyxf −−= (level curve = 0, 1, 2, 3)

Limit and Continuity 4. Find the limit, if exists, or show that the limit does not exist.

(a). 2 2

2 2( , ) (0,0)

sinlim2x y

x yx y→

++

[limit does not exist]

(b). 2 2

2 2( , ) (0,0)lim

1 1x y

x yx y→

+

+ + − [2]

(c). [1] ( , , ) (3,0,1)

lim sin( / 2)xy

x y ze zπ−

→

(d). 3

2( , ) (0,0)lim

x y 2

xy yx y→

++

[limit does not exist]

(e) ( ) ( ) 44

4

0,0, 3lim

yxy

yx +→ along (i) y = 2x and (ii) y = x2. (Sept 2012)

[limit does not exist]

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5. Determine the set of points of domain at which the function is continuous:

(a) 2

sin( )( , ) x

xyf x ye y

=−

[ ] / 2{( , ) | }xx y y e≠ ±

(b). [ ] 2 2( , ) ln( 4)f x y x y= + − 2 2{( , ) | 4}x y x y+ > (c). 1 2 2( , ) sin ( )f x y x y−= + [ ] 2 2{( , ) | 1}x y x y+ ≤ Partial Derivatives

6. Find the indicated partial derivatives. (a). 2 3

,( , , ) ln( ), rss rstf r s t r rs t f f= [ ] 22 / , 0s−

(b). 3

, zz u v wu v w∂

= −∂ ∂ ∂

[ ( ) 2/3

41 −− wv ]

(c) Let 2222 122),(

yxyxyxf ++= . Find 2

2

xf

∂

∂ , y

f2

2

∂

∂ and yxf2

∂∂∂ .

(exam Sept 2010) [ .4;64;64 33

2

422

2

242

2

yxyxf

yxyf

yxxf

=∂∂

∂+=

∂∂

+=∂∂

]

7. (i) Verify that is a solution of the heat conduction

equation

2 2exp( )sin( )u k tα= − kx

x

2

2 .t xu uα=

(ii) Verify that the following function satisfies this condition: fxy = fyx. f(x, y) = x sin y + y sin x + xy (Exam Sept 2013)

8. Verify that the function is a solution to the three- 2 2 2 1/( , , ) ( )u x y z x y z −= + + dimensional Laplace equation 0.xx yy zzu u u+ + =

9. The temperature at a point (x,y) on a flat metal plate is given by

2 2

60( , )1

T x yx y

=+ +

,

where T is measured in °C and x, y in meters. Find the rate of change of temperature with respect to distance to the point (2, 1) in (a). the x-direction; and [ ] 0( 20 / 3) /C m− (b). the y-direction. [ ] 0( 10 / 3) /C m−

2

10. The gas law for a fixed mass m of an ideal gas at absolute temperature T , pressure P and volume V is PV=mRT, where R is the gas constant. Show that

(a). 1P V TV T P∂ ∂ ∂

= −∂ ∂ ∂

(b). P VT mT T∂ ∂

=∂ ∂

R

Tangent Plane and Linear Approximation, Differentials 11. Find an equation of the tangent plane to the given surface at the specified point: (a). 2 24 2 , (1, 1,1)z x y= − − − [ ] 2 4x y z− + =

(b). [ln , (1, 4,0)z y x= 4 4z x= − ] (c). [ ] 2 2exp( ), (1, 1,1)z x y= − − 2 2z x y= + +1(d) f(x, y, z) = cos πx – x2y + exz + yz at (0,1,2) [f(x, y, z)=2x + 2y + z]

12. (a). Find the linear approximation of the function ( , ) ln( 3 )f x y x y= − at (7,2)

and use it to approximate f(6.9,2.06). [-0.28] (b). Find the linear approximation of the function 2 2( , , ) 2f x y z x y z= + + at

(3,2,6) and use it to approximate the number 2 2(3.02) (1.97) (5.99) .+ + 2

2

[6.9914]

(c). If and (x,y) changes from (3,-1) to (2.96,-0.95), compare 2 3z x xy y= − + the value of and dz. [zΔ 0.73, 0.7189dz z= − Δ = − ] (d) The kinetic energy, T is related to the mass, m and speed, v as given by

.2

2mvT = Find the approximate change in energy, ΔT, if the mass, m changes

from 50 to 50.5, and speed, v changes from 100 to 110.(Exam Feb 2013) [52500]

13. Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. [ ] 38.8cm

14. Given a velocity function ( ) 22 3, tststsV −+= , where s is displacement and t is time. If s changes from 2 to 2.05 and t changes from 3 to 2.96, find the approximate change in V using differentials. (exam Sept 2010) [0.65]

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Chain Rule and Implicit Differentiation

15. Use the chain rule to find the indicated partial derivatives: 2 2 2( ). ln( ), 2 , 2 , 2 ;

, 1.

a R u v w u x y v x y w xyR R when x yx y

= + + = + = − =∂ ∂

= =∂ ∂

[ 2 2 2

2 4 4u v wyu v w+ ++ +

; 97

]

[ 2 2 2

4 2 4u v wxu v w− ++ +

, 97

]

2( ). , cos , sin ,

, , 2, 3, 0

b u x yz x pr y pr z p ru u u when p rp r

θ θ

θθ

= + = = = +∂ ∂ ∂

= = =∂ ∂ ∂

2 cos sin ,36;2 cos sin ,24;

2 sin cos , 30

xr zr yxp zp y

xpr zpr

θ θθ θθ θ

+ +⎡ ⎤⎢ ⎥+ +⎢ ⎥⎢ ⎥− +⎣ ⎦

(c) If u = yx2 + xy2 , where x = s + 3t and y = 2s – t, determine .tu∂∂

[ ] 2234 xyxy −+

16. Use the implicit differentiation to find zx∂∂

and zy∂∂

:

(a). cos( )xyz x y z= + + [ sin( )sin( )

xz x y zxy x y z+ + +

−+ + +

]

(b). arctan( )x z− = yz [ 2 21z

y y z−

+ +]

17. Given , find 9742 42 =+− yxyx dxdy

using implicit differentiation.

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

xyxy

37

18. If u = f(x,y), where x = es cos t, y = es sin t, show that

22 2

2su u u uex y s t

−2⎡ ⎤⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = +⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦

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Critical (Stationary) Points 19. Find the local maximum and minimum values and saddle points of the function. (a) . [(0,0) local max.; (0,2) local min; 2 3 2 2( , ) 3 3 3 2f x y x y y x y= + − − +

( 1,1)± saddle points]

(b) 2 2

( , ) x yf x y xye− −= . [(0,0) saddle pt.);1 1( ,2 2

± m ) local min;

1 1( local max.] , )2 2

± ±

(c) f(x, y) = (4x-1)2 + (2y + 4)2 + 1 (Exam May 2013) [(1/4, -2) local min.] Lagrange Multipliers 20. Use Lagrange multipliers to find the maximum and minimum values of the function

subject to the given constraint. (a) ; . [max f(2,3)=26; min. f(-2,-3)=-26] yxyxf 64),( += 1322 =+ yx(b) ; . [max. 22),( yxyxf += 144 =+ yx 2 , min1]

(c) ; . [max.xyzzyxf =),,( 632 222 =++ zyx 2 / , min. .3 2 / 3−

21. The production output P of a small factory with x workers and y units of equipment is estimated to be . Due to limitations in factory space and available capital, the factory manager has no choice but to impose the constraint

( ) 5/35/250, yxyxP =

200=+ yx . Find the maximum possible production output by applying the method of Lagrange multipliers. (Exam Sept 2010) [5101.698 units]

22. One children products of the factory with x workers and y units of equipment has production output as P(x, y) = xy. Due to fluctuant on demands, the factory manager has imposed the following constraint: x2 + y2 – xy = 4. Find the maximum possible production output by applying method of Lagrange multipliers. [4 units]

23. Use Lagrange multipliers to find the maximum and minimum values of the function f(x, y) = x2y subject to the constraint g(x, y) = x2 + 2y2 – 6. (Exam Sept 2012) [max. 4; min -4] 24. The production output F of a factory with x workers, y units of equipment and z units of materials are estimated as F(x, y, z)= 2x+ z2. Due to limitations in factory conditions, the factory has imposed the constraint of x2 + y2 + 2z2 = 25. Find the maximum possible production output by applying the method of Lagrange multipliers. (Feb 2014)

( ) ]229

221,0,2,[ unitsPyxF =⎟⎟

⎠

⎞⎜⎜⎝

⎛±= ……End

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