Semester MATH1052 2 - 2008 Solutions to Problem Sheet 6
1. Since V = IR = R2 . So I IR1 R2 = it follows that V is a function of I, R1 and 1 1 R1 + R2 + R1 R2
dV V dI V dR1 V dR2 = + + dt I dt R1 dt R2 dt Now V R1 R2 = I R1 + R2 V = IR2 R1 V = IR1 R2 So 1(R1 + R2 ) R1 (R1 + R2 )2 1(R1 + R2 ) R2 (R1 + R2 )2 = 15 8 =
2 IR2 (R1 + R2 )2 2 IR1 . (R1 + R2 )2
=
V (2, 3, 5) I
V 2.52 50 (2, 3, 5) = = 2 2 R1 (3 + 5) 8 V 2.9 18 (2, 3, 5) = = 2. 2 R2 8 8 Now to put it all together dV 15 1 50 1 18 1 = + 2 + 2 = 0.38125 dt 8 100 8 2 8 10 amps ohms/sec.
2) a) i) Since u(x, y, z) = xy + yz + zx, where x(s, t) = st, y(s, t) = est , z(s, t) = t2 u x u y u z u = + + s x s y s z s Now So u u u = y + z, = x + z, = y + x. x y z
u = (est + t2 )t + (st + t2 )test + (est + st). s And evaluated at s = 0, t = 1. u s = (1 + 1).1 + (0 + 1).1 = 3 1
(0,1)
ii) If z(x, y) = then
x , where x(r, s, t) = rest and y(r, s, t) = rset y z z x z y = + r x r y r
Now So
z 1 z = , = x y y
x y2
,
x y = est and = set . r r z et est rest set = r rs (rset )2 z r 1 1 1 = . 2 4 4 1 4
at
(1,2,0)
=
Similarily z s and z t
(1,2,0) (1,2,0)
=
et rtest rest ret rs (rset )2
(1,2,0)
=
= 1 . b) Dierentiate 2
d(xcos(y) + ycos(x)) d(1) = = 0. dx dx Then thinking of y as a function of x, y(x) and using the chain and product rules cosy xsiny dy dy + cosx ysinx = 0 dx dx
dy This implies that (cosx xsiny) = ysinx cosy dx So that dy ysinx cosy = . dx cosx xsiny c) Stewart 6th Ed - Section 15.5 Q38 The volume of a right circular cone is r2 h V = , where r is the base radius and h is the height. 3 We are given that dh dr = 1.8 in/s and = 2.5 in/s. when r = 120 ins and h = 140 ins. dt dt By the chain rule dV V dr V dh = (120, 140) + (120, 140) dt r dt h dt 2rh V (120, 140) = r 3 V r2 (120, 140) = h 3 2
. Now
(120,140)
= 11200
and
(120,140)
= 4800
Finally 3)
dV = 11200.1.8 + 4800.(2.5) = 8160. dt
i) a) First method: dw w dx w dy w w = + ; since = y, =x dt x dt y dt x y dx dy dx dy = y +x since = 2 cos t, = sin t dt dt dt dt = cos t.2 cos t + 2 sin t( sin t) = 2(cos2 t sin2 t) = 2 cos 2t.
(b) Second method: w = 2 sin t cos t implies that: dw = 2 cos t cos t 2 sin t sin t = 2(cos2 t sin2 t) = 2 cos 2t. dt ii) (a) First method: Note that z = arctan t implies that tan z = t dz = 1. and sec2 z dt dw w dx w dy w dz = + + dt x dt y dt z dt dx dy dz = y tan z + x tan z xy sec2 z dt dt dt = t2 .t.1 + t.t.2t + t.t2 .1 = 4t3 (b) Second method: w = t3 .t = t4 iii) (a) First method: dw dx dy dz = yz + xz + xy dt dt dt dt = 2t.et .2t + t2 et .2 + t2 .2t(et ) = (6t2 2t3 )et . (b) Second method: w = 2t3 et 4. a) dw = 6t2 et 2t3 et = (6t2 2t3 )et . dt dw = 4t3 . dt
w w r w = + . Now x = r cos , y = r sin , r = x2 + y 2 and x r x x r x y sin x y = sec2 = 2 = and = = cos , tan = . So 2 + y2 x x r x x r cos2 x sin = . x r w w w sin So = cos . x r r w r w r y 1 cos w = + , = = sin and = = . So b) 2 y r y y y r y x sec r w w cos w = sin + . y r r 3
c) w x
2
+
w y
2
= = =
w w sin cos r r w r w r
2 2 2
2
+
w w cos sin + r r w
2
2
(cos + sin ) +
2
(cos2 + sin2 ) r2
+
w
2
1 r2
5.
a) f = (1 2x2 )ye(x
2 +y 2 )
i + (1 2y 2 )xe(x
2 +y 2 )
j
So if f = 0 then (1 2x2 )y = 0 and (1 2y 2 )x = 0. 1 Now if (1 2x2 )y = 0 then y = 0 or x = 2 . If y = 0 then (1 2y 2 )x = 0 implies that x = 0 and 1 1 if x = 2 then (12y 2 )x = implies 0 that y = 2 . So the critical points are 1 1 1 1 1 1 1 1 (0, 0), ( , ), ( , ), ( , ), ( , ) 2 2 2 2 2 2 2 2 Now 2f 2 x2 y 2 , 2 = 2x(3 2x )ye x 2f 2 x2 y 2 , 2 = 2y(3 2y )xe y
2f 2 2 = (1 2x2 )(1 2y 2 )ex y . yx and D = 2f 2f 2f 2 ( ) so at (0, 0) D(0, 0) = 1 < 0 which implies a xy x2 y 2
saddle. 1 1 1 1 Now at ( 2 , 2 ) and at ( 2 , 2 ) D=
2(3 1)(3 1)e1 >0 4
2f and < 0, which implies a local max. x2 1 1 1 1 Now at ( 2 , 2 ) and at ( 2 , 2 ) 2f D > 0 and > 0, which implies a local min. x2 b) f = (8y (x+y)3 )i+(8x(x+y)3 )j. So if f = 0 then 8y = (x+y)3 = 8x. This implies x = y. So 8x3 = 8x. Therefore, x = 0 or x = 1. Moreover, 2f = 3(x + y)2 , x2 Now at (0, 0) 2f = 0, x2 2f = 8, xy 4 and 2f = 0. y 2 2f = 8 3(x + y)2 , xy and 2f = 3(x + y)2 . y 2
This leads to D < 0. So the point (0, 0) is a saddle point. At (1, 1) or (1, 1) 2f = 3.4 = 12, x2 2f = 8 3.4 = 4, xy and 2f = 3.4 = 12. y 2
This leads to D = 122 42 > 0. Since (1, 1) are both local maxima. c)
2f = 12 < 0 the points (1, 1) and x2
f = cos x sin yi + sin x cos yj. So if f = 0 then either cos x = cos y = 0 or sin x = sin y = 0. This implies either x = n + /2 and y = m + /2 or x = n and y = m. Moreover, 2f = sin x sin y, x2 2f = cos x cos y, xy and 2f = sin x sin y. y 2
Now at x = n + /2 and y = m + /2 2f = 1, x2 2f = 0, xy and 2f = 1. y 2
This leads to D = 1 > 0. Now depends on n or m to be even or odd we will nd minimum or maximum points. Now at x = n and y = m 2f = 0, x2 2f = 1, xy and 2f = 0. y 2
This leads to D = 1 < 0. So the points x = n and y = m are saddle points. d) f = ex (1 cos y)i + ex sin yj. So if f = 0 then cos y = 1 and sin y = 0. This implies y = 0, 2, 4, . . . but x is unspecied. Moreover, 2f x 2 = e (1 cos y), x At y = 0, 2, 4, . . . we see 2f = ex sin y, xy and 2f x 2 = e cos y. y
2f 2f 2f x = 0, = 0, and 2 2 = e . This xy x y implies D = 0. In fact there is a minimum at y = 0, 2, 4, . . . . But it is not a local min as it occu...