MA261-A Calculus III 2006 Fall Homework 7 Solutions Due 10/20/2006 8:00AM

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1 MA26-A Calculus III 2006 Fall Homework 7 Solutions Due 0/20/2006 8:00AM 3 #4 Find the rst partial derivatives of the function f (; ) f (; ) f (; ) #6 Find the rst partial derivatives of the function z ln ln 3 #24 Find the rst partial derivatives of the function f (; ) Z cos t 2 Note that b the Fundemental Theorem of Calculus, when a is a constant, the derivative of R f (t) is f () a f (; ) cos 2 f (; ) d d Z cos t 2 cos 2

2 2 3 #26 Find the rst partial derivatives of the function f (; ; z) 2 e z There are three rst partial derivatives f (; ; z) 2e z, f (; ; z) 2 ze z, f z (; ; z) 2 e z 3 #36 Find the partial derivative f ( 6; 4) where f (; ) sin (2 + 3) We have f (; ) cos (2 + 3) (3) 3 cos (2 + 3) Thus, 3 #42 Use implicit di erentiation to nd f ( 6; 4) 3 cos (2 ( 6) + 3 (4)) 3 where z ln ( + z) Treat as a constant Then take the partial derivative with respect to for the both sides of the equation We get This implies that or, + z +z +z 3 #46 Find of the function (a) z f () g () (b) z f () (c) z f + + z + z, +z (+z) +z (a) f () g () f () g () (b) [f ()] [f ()] (c) h i f h i f 2 ( + z),

3 3 3 #48 Find the second partial derivatives of the function f (; ) ln (3 + 5) f (; ) f (; ) So, there are four second partial derivatives f (; ) f (; ) f (; ) f (; ) d 3 (3 + 5) d d 3 d (3) (5) 3 d (3 + 5) d (3 + 5) 2 (3 + 5) 3 d (3 + 5) d (3 + 5) 2 d 5 (3 + 5) d d 5 d 5 d (3 + 5) d (3 + 5) 2 (3 + 5) 5 d (3 + 5) d (3 + 5) 2 4 #2 Find an equation of the tangent plane to the given surface z e 2 2 at the speci ed point (; ; ) e2 e2 2 2 (2) ( 2) 9 (3 + 5) 2, 5 (3 + 5) 2, 5 (3 + 5) 2, 25 (3 + 5) 2 At the point (; ; ), we have ( e()2 ) 2 (2 ()) 2 (; ) e ()2 ( ) 2 ( 2 ( )) 2 Thus, the equation of the tangent plane at the point (; ; ) is z (; ) ( ) + (; ) ( ( )) 2 ( ) + 2 ( + )

4 4 4 #0 Eplain wh the function f (; ) is di erentiable at the point (6; 3) f (; ) f (; ) 2 At the point (6; 3), both f (6; 3) 3 f (6; 3) eist Moreover, both f f are continuous since the are de ned rational functions B the theorem, f (; ) is di erentiable at the point (6; 3) 4 #34 Find an equation of the tangent plane to the given parametric surface u 2, v 2, z uv, to the point (u; v) (; ) If ou have software that graphs parametric surfaces, use a computer to graph the surface the tangent plane We write r (u; v) (u; v) i + (u; v) j + z (u; v) k u 2 i + v 2 j + uvk r u (u; v) 2ui + 0j + vk 2ui + vk The normal vector is i j k r u r v 2u 0 v 0 2v u r v (u; v) 0i + 2vj + uk 2vj + uk 0 v 2v u 2v 2 i 2u 2 j + 4uvk i 2u v 0 u j + 2u 0 0 2v k At the point (u; v) (; ), the (; ; z) (; ; ) r u r v (; ) ( 2; 2; 4) So, the equation of the tangent plane is The graph is 2 ( ) 2 ( ) + 4 (z ) 0

5 z #36 Find an equation of the tangent plane to the given parametric surface r (u; v) uvi + u sin vj + v cos uk to the point (u; v) (0; ) If ou have software that graphs parametric surfaces, use a computer to graph the surface the tangent plane r u (u; v) vi + sin vj At the point (u; v) (0; ), we have v sin uk r v (u; v) ui + u cos vj + cos uk r (0; ) (0) () i + (0) sin () j + () cos (0) k k, r u (0; ) () i + sin () j () sin (0) k i, r v (0; ) (0) i + (0) cos () j + cos (0) k k The normal vector at the point (u; v) (0; ) is i j k r u r v (0; ) i 0 0 j k j So, the equation of the tangent plane is 0 ( 0) ( 0) + 0 (z ) 0, or, 0

6 6 4 #38 Suppose ou need to know an equation of the tangent plane to a surface S at the point P (2; ; 3) You don t have an equation for S but ou know that the curves r (t) h2 + 3t; t 2 ; 3 4t + t 2 i r 2 (u) h + u 2 ; 2u 3 ; 2u + i both lie on S Find an equation of the tangent plane at P From these two curves, we have r 0 (t) h3; 2t; 4 + 2ti r 0 2 (u) h2u; 6u 2 ; 2i The point P (2; ; 3) lies in r when t 0 Also, the point P (2; ; 3) lies in r 2 when u Thus, at the point P (2; ; 3), r 0 (0) h3; 0; 4i r 0 2 () h2; 6; 2i are two tangent vectors Therefore, the normal vector of the tangent plane is i j k r 0 (0) r 0 2 () i j k 24i 4j + 8k So, the equation of the tangent plane is 4 #40 Show that the function 24 ( 2) 4 ( ) + 8 (z 3) 0 f (; ) 5 2 is di erentiable b nding values of 2 that satisf De nition 7 First we have f (; ) The increment of z at (a; b) is f (; ) 0 z f (a + ; b + ) f (a; b) (a + ) (b + ) 5 (b + ) 2 ab 5b 2 ab + () b + a () + () () 5b 2 0b () + 5 () 2 ab + 5b 2 () b + a () + () () 0b () + 5 () 2 b () + a () 0b () + () () + 5 () 2 b () + (a 0b) () + () () + 5 () () If we set 2 5, then at (a; b), we have z f (a; b) () + f (a; b) () + () + 2 (), where 2! 0 as (; )! (0; 0) Therefore, f is di erentiable at (a; b)

7 4 #42 (a) The function f (; ) if (; ) 6 (0; 0) 0 if (; ) (0; 0) was graphed in Figure 4 Show that f (0; 0) f (0; 0) both eists but f is not di erentiable at (0; 0) (b) Eplain wh f f are not continuous at (0; 0) (a) B the de nition, f (0 + h; 0) f (0; 0) f (0; 0) lim h!0 h f (0; 0) lim h!0 f (0; 0 + h) f (0; 0) h Thus, f (0; 0) f (0; 0) both eists Consider lim h!0 lim h!0 z f (0 + ; 0 + ) f (0; 0) h0 0 h h 0h h 2 h lim h!0 0 0 lim h!0 0 0 () () () 2 + () 2 To see the di erentiabilit, we need to nd 2 such that z f (a; b) () + f (a; b) () + () + 2 () 0 () + 0 () + () + 2 () () + 2 () where 2! 0 as (; )! (0; 0) But, () 2 +() 2 goes to as (; )! (0; 0) Thus, 2 cannot be found Therefore, f is not di erentiable at (0; 0) (b) To see if f f are continuous at (0; 0), we need to check if lim f (; ) f (0; 0) (;)!(0;0) lim f (; ) f (0; 0) (;)!(0;0) (2 + 2 ) (2) ( ) ( ) 2 (2 2 ) f (; ) ( ( 2 2 ) ( ) 2 if (; ) 6 (0; 0) 0 if (; ) (0; 0) Consider lim f ( 2 2 ) (; ) lim (;)!(0;0) (;)!(0;0) ( ) 2 When we approach along 0, the limit becomes ( ) 2, we have ( ) lim (0;)!(0;0) ( ) 2 lim 3 (0;)!(0;0) lim 4 (0;)!(0;0) The limit does not eist When we approach along 0, the limit becomes (0 2 2 ) 0 lim (;0)!(0;0) ( ) 2 lim 0 0 (0;)!(0;0) 7

8 8 Therefore, the limit lim (;)!(0;0) f (; ) does not eist So, it is not equal to f (0; 0) This tells us that f is not continuous at (0; 0) Similarl, Consider f (; ) (2 + 2 ) (2) ( ) ( ) 2 (2 2 ) ( ) 2, we have ( ( 2 2 ) ( ) 2 if (; ) 6 (0; 0) 0 if (; ) (0; 0) ( 2 2 ) lim f (; ) lim (;)!(0;0) (;)!(0;0) ( ) 2 When we approach along 0, the limit becomes (0 2 2 ) lim (0;)!(0;0) ( ) 2 lim 3 (0;)!(0;0) lim 4 (0;)!(0;0) The limit does not eist When we approach along 0, the limit becomes ( ) 0 lim (;0)!(0;0) ( ) 2 lim 0 0 (0;)!(0;0) Therefore, the limit lim (;)!(0;0) f (; ) does not eist So, it is not equal to f (0; 0) This tells us that f is not continuous at (0; 0) 5 #4 Use the Chain Rule to nd dw B the Chain Rule, we have where w + z 2, e t, e t sin t, z e t cos t d d dz () e t + + z 2 e t sin t + e t cos t + (2z) e t cos t e t sin t e t sin t e t + e t + e t cos t 2 e t sin t + e t cos t + 2 e t sin t e t cos t e t cos t e t sin t 2e 2t sin t + e 2t cos t + e 3t (sin t + cos t) + 2e 3t sin t cos t (sin t cos t) 5 #8 Use the Chain Rule to nd dz ds dz where z sin tan, 3s + t, s t

9 9 B the Chain Rule, we have dz @s (cos tan ) (3) + sin sec 2 () 3 cos tan + sin sec 2 3 cos (3s + t) tan (s t) + sin (3s + t) sec 2 @t (cos tan ) () + sin sec 2 ( ) cos tan sin sec 2 cos (3s + t) tan (s t) sin (3s + t) sec 2 (s t) 5 #8 Use the Chain Rule to nd the partial derivatives du, du d d where du u p r 2 + s 2, r + cos t, s + sin t when, 2, t 0 Note that when, 2, t 0, we have r 2+cos 0 3 s +2sin 0 This tells us that u p r 2 + s 2 p p 0 B the Chain Rule, when, 2, t 0, we have @s r s p (cos t) + p () r2 + s 2 r2 + s 2 3 p (cos 0) + p () p 0, @s r s p () + p (sin t) r2 + s 2 r2 + s 2 3 p () + p (sin 0) p 0

10 @t r s p ( sin t) + p ( cos t) r2 + s 2 r2 + s 2 3 p ( sin 0) + p (2 cos 0) p 0 5 #22 Use equation 6 to nd d d where Let e 2 F (; ) e 2 We have 23 2e e 2 Thus, b equation 6, we have 5 #26 Use equation 7 to nd Let d d where 2 3 2e e 2 z cos ( + + z) F (; ; z) cos ( + + z) We have sin ( + + z) z, sin ( + + z) z, sin ( + + z) Thus, b equation 7, we have dz d sin ( + + z) z sin ( + + z) z sin ( + + z) + z sin ( + + z) +

11 dz d sin ( + + z) z sin ( + + z) + z sin ( + + z) sin ( + + z) + 5 #34 The voltage V in a simple electrical circuit is slowl decreasing as the batter wears out The resistance R is slowl increasing as the resistor heats up Use Ohm s Law, V IR, to nd how the current I is changing at the moment when R 400, I 0:08A, dv 0:0V/s, dr 0:03/s We have Thus, This tells us that dv di R + I dr 0:0 di 0:0 0:08 0: :08 0:03 di 0: #40 If z f (; ), where s + t s t, shwo that B the Chain Rule,