aa midterm solutions engr 233

7/26/2019 AA Midterm Solutions ENGR 233

1/5

ENGR-233: Applied Advanced Calculus

Midterm sample 1 solutions

Problem 1.Find the parametric equation of the line of intersection of two

planes:

1 : x+y8z=4 and 2: 3xy+4z=0 .

olution: Let us eliminate the variables x and y from the equations.

ddin! the equations for 1 and 2 we !et the equation 4x4z=4 " orxz=1 # so" x=z+1 . $ubstitutin! this into the equation for 2 we !ety=3x+4z=3z+3+4z=%z+3 . $o" we have e&pressed both x and y in

terms of z . 'ence" the parametric equation of the line of intersection of

P1 and P2 are: x=t+1" y=% t+3" z=t .

Problem 2.(osition vector of a movin! particle is !iven b)

r(t)= 3 t2+1" 2 t2% t+3"(t1)2) .*a+ t what time*s+ does the particle pass the xz ,plane-

*b+ hat are the particle *i+ coordinates" *ii+ velocit)" *iii+ speed" *iv+

acceleration at t=2 -

olution: *a+ e have to find t such that y=2 t2% t+3=0 # so" we haveto solve the quadratic equation 2 t

2% t+3=0 . /ts solution:

t=%424

4 =

%4

# t1=1 /2" t2=3 .

*b+ r(t)=(3 t2+1" 2 t2% t+1"(t1)2) #

r( t)=( t , 4 t%"2 t2) # r( t)=("4"2) .

r(2)=(13"3"1) # r(2)=(12"1"2) # r(2)=14 # r(2)=("4"2) .


2/5

Problem 3.Find the directional derivative of F(x , y , z)=1x2 ez+3y2 inthe direction u=(4"4"2) at the point (1"2"0) .

olution. fx

=30x ez # F

y=y #

F

z=1x2 ez .

$o"Fx

(1"2"0)=30 "Fy

(1"2"0)=12 "Fz

(1"2"0)=1 .

ow" u=1+1+4=3= # so" the normalied vector

v=

u

u=(2 /3"2/3"1/3) . 5hen"

D uF=Fv=302

3+12

23 +(1)

1

3=% .

(roblem 4. Let F=(x (x2+y2+z2)m, y (x2+y2+z2)m, z(x2+y2+z2)m) .*a+ Find F # *b+ Find m such that F=0 for x2+y2+x2>0 .

olution. F=Px

+Qy

+Rz

=x

(x (x2+y2+z2)m)+ y

(y (x2+y2+z2)m)+ z

(z(x2+y2+z2)m)

=(x2+y2+z2)m+xm(x2+y 2++z2)m12x+(x2+y2+z2)m+ym(x2+y2+z2)m12y

+(x2+y2+z2)m+zm(x2+y2+z2)m12z=3(x2+y2+z2)m+2m(x2+y2+z2)(x2+y2+z2)m1

=(3+2m)(x2+y2+z2)m .

*b+ F=0 if 3+2m=0 " i.e. m=3 /2 .


3/5

Problem !.Let

F(x , y , z)=(a cosy+bsinz , c cosz+dsinx , ecosx+ fsiny ) .

*a+ Find F # *b+ Find the values of a , b , c , d , e , f such thatFF .

olution.

*a+ F= i j k

x

y

z

a cosy+bsinz c cosz+dsinx e cosx+fsiny

=(f cosy+csinz , b cosz+e sinx , dcos x+a siny ) .

*b+ 6omparin! the coefficients we conclude that FF ifa=f , b=c , e=d .

Problem ".Find the wor7 done b) the force F(x , y , z)=(xy , x2,z)movin! a particle alon! a line se#mentfrom a point P(1"2"3) to a point

Q (2"1"2) .$int: Find the parametric equation of the line connectin! P and Q " then

evaluate the inte!ral.

olution. 5he parametric equations of the se!ment connectin! P and Q is

r(t)=(1+t ,1t ,3t) *09t91+. 5hen r'(t)=(1"2"1) . 5hen thewor7 done b) the force F on the se!ment is

W=0

1

(2t,(1t)2"3+t)(1"1"1)dt=0

1

(2 t12 tt2+3t)dt

=13

1

2+2=

%

.


4/5

Problem %.Let F(x , y , z)=(y exy, x exysin(y+z),3z2sin (y+z)) .*a+ $how that

C

Fdr is independent of the path#

*b+ 6ompute the inte!ral for an) path C from the point A(2"1"1) to thepoint B(3"2"2) .

olution. Let us chec7 the conditions for the path independence.

Qx

Py

=exy+xyexyexyyxexy0 #

R

x

P

z=000 #

Ry

Qz=cos (y+z)+cos(y+z)0 .

'ence" there e&ists a function (x , y , z) such that F= " i.e.

x=P=yexy "

y

=Q=xexysin (y+z) "

z

=R=3z2sin (y+z) .

From the first equation we have=yexy dx=y1

ye

xy+g(y , z)=exy+g(y , z);

5hen"

y

=xexy+gy

xexysin (y+z) ;


5/5

'ence

gy

=sin (y+z) ; g= (sin(y+z))dy=cos(y+z)+h (z) #

5hen"

=exy+cos(y+z)+h(z) #z

=sin (y+z)+h '(x)=32sin (y+z);

h '(z)=3z2 ; h (z)=z3+C ; (x , y , z)=exy+cos(y+z)+z3 . ou can

chec7 )ourself thatx

=P , y

=Q , z

=R .

5hen"

A

B

Fdr=(B)(A)=(3"2"2)(2"1"1)=e+cos(0)8e21+1

=ee2 .

aa midterm solutions engr 233

Documents