MTH301 Midterm Past Paper Spring 2006 Set 2

VUSR · MTh301-Calculus II · Midterm Special 2006

?� At critical points function can assume o maximum value. o minimum value. o both maximum and minimum value.

o zero value

?� A plane can be perfectly determined by o one point. o one point and normal vector. o one point and parallel vector. o one normal vector.

?� The length and width of a rectangle are measured with the errors of at most 4 % and 3%, respectively. Use differentials to approximate the maximum percentage error in the calculated area.

Solution:

Let x = length

y = width

A = area

Then A xy So

?�A=y ?�x

?�A=x ?�y

Therefore,

?�A ?�A ?� ˜�A ?�y ?�x ?�y

y x x y

VUSR

We desire percentage change in A, which is relative change multiplied by 100 so let’s work out relative change first. This is given by

A y x+x y A A A ?�x ?�y

x y Since

A xy

?�x 0.04

1. 04 =� x ?�y 0.04 and 0.03 =� y

The maximum possible value for ?�A given the above constraints. This happens A when ?�x=0.04 and ?�y=0.03, giving ?�A=0.07 . This is relative error, so the x y A percentage error is 7%.

Convert the Cartesian coordinates (1, 2 , 2) -� into cylindrical coordinates.

Solution:

We are to convert (x,y,z) int (r, ?� ,z)

Given:

x = 1

y = 2

z = -�2

We know that:

r = x 2+ y

Put value of x and y in the above equation: