MATH 1014.03MW Exam 1 Monday, January, 24 2000

NAME:

Student Number:tex2html_wrap_inline266 No.tex2html_wrap_inline268Marks tex2html_wrap_inline270


Instructions: This exam contains 4 questions and has a total of 100 marks. Show all of your work.

tex2html_wrap_inline274(30 Marks)A 5 meters long ladder is leaning against a building. The bottom of the ladder is dragged along the ground, away from the building, at 3 meters per minute. How fast is the top of the ladder moving down the side of the building when it is 3 meters above the ground?

Solution (i). We draw the following figure.

1.gif (2102 bytes)

(ii). Hypotheses: x'(t)=3, |AB|=5 and tex2html_wrap_inline290.

(iii). Question: Find y'(t) when y(t)=3.

(Method 1) By the above figure, we have
 equation33

Taking the derivative of Eq. tex2html_wrap_inline300 relative to t, we have
displaymath304

This implies
 equation40

By Eq. tex2html_wrap_inline306, we have tex2html_wrap_inline308. This, together with Eq. tex2html_wrap_inline310, implies
displaymath312

When x'(t)=3 and y(t)=3, we have tex2html_wrap_inline318.

(Method 2) By the above figure, we have
 equation54

This implies
 equation60

Taking the derivative of Eq. tex2html_wrap_inline322 relative to t, we have
 equation65

By Eq. tex2html_wrap_inline326, we have tex2html_wrap_inline308. This, together with Eq. tex2html_wrap_inline330, implies
displaymath312

When x'(t)=3 and y(t)=3, we have tex2html_wrap_inline318.

(Method 3) By the above figure, we have
 equation81

This implies
 equation87

Since x'(t)=3, x(t)=3t. This, together with Eq. tex2html_wrap_inline346, implies
 equation92

Taking the derivative of Eq. tex2html_wrap_inline348 relative to t, we have
 equation98

When y(t)=3, by Eq. tex2html_wrap_inline354, we have tex2html_wrap_inline356. Since x(t)=3t, t=x(t)/3=4/3. Hence, when y(t)=3, that is, t=4/3, we have
displaymath366

tex2html_wrap_inline368(25 Marks) Find the area of the region enclosed between the graphs of the functions
displaymath370

Solution: Let tex2html_wrap_inline372. Then tex2html_wrap_inline374.

 

a1014.gif (3728 bytes)

Hence, x=-2, x=0 and x=2. We draw the following graph.


eqnarray118


eqnarray132

Hence, the area of the region =|A|+|B|=4+4=8.

tex2html_wrap_inline384(25 Marks)Find the volume of the solid generated by revolving the region bounded by the parabolas tex2html_wrap_inline386 and tex2html_wrap_inline388 about the x-axis.

Solution: Let tex2html_wrap_inline392. Then x=0 and x=1. We draw the following graph.

b1014.gif (2753 bytes)

 

(Method 1)

Region: tex2html_wrap_inline386 and tex2html_wrap_inline402 for tex2html_wrap_inline404.

Axis of revolution: x-axis.

The volume tex2html_wrap_inline408.

(Method 2) Region: tex2html_wrap_inline412 and tex2html_wrap_inline414, tex2html_wrap_inline416.

Axis of revolution: x-axis.

The volume tex2html_wrap_inline420.

tex2html_wrap_inline422 (20 Marks) Write a sum of two integrals which equals the area of the region inside tex2html_wrap_inline424 and tex2html_wrap_inline426.

(You need not evaluate the integrals).

Solution: (i) We draw the following graph.

c1014.gif (3522 bytes)

 

(ii) Find the points of intersection.

Let tex2html_wrap_inline432. Then tex2html_wrap_inline434 This implies
displaymath436

Hence, tex2html_wrap_inline438 or tex2html_wrap_inline440. So, we obtain tex2html_wrap_inline442 or tex2html_wrap_inline444.

(iii) Using the formulas of the areas of polar equations, we have
displaymath448

Kunquan Lan
Mon Jan 24 11:39:17 EST 2000