## Some answers to the secret review problems for the final exam in Math 151:04-06, fall 2003

Solutions
D E P Q S2
a, b

Problem D
Particle A moves along the positive x-axis and particle B moves along the positive y-axis.
a) If $A$ is at the point (a,0) and B is at the point (0,b), write an expression for the distance from A to B.

Solution

Expression for distance between A and B:
Pythagorean Theorem gives...
d^2 = a^2 + b^2
d = (a^2 + b^2)^(1/2)

b) At a certain time, A is at the point (7,0) and moving away from the origin with a speed of 4 units/sec. At the same time B is at the point (0,5) and moving towards the origin at 2 units/sec. At what rate is the distance from A to B changing at that time? Are the particles moving towards each other or moving away from each other at that time?

Solution

Given: da/dt=-2 and db/dt=(t)=4
Chain rule...
dd/dt=(1/2)[(a^2+b^2)^(-1/2)](2a(da/dt)+2b(db/dt))
a=7, b=5, a'=-2, b'=4
dd/dt=(1/2)[(7^2+5^2)^(-1/2)](2(7)(-2)+2(5)(4))
dd/dt=(1/2)[(74)^(-1/2)](12)
dd/dt=6*sqrt(74)
This value is positive so the particles are moving away from each other.

Submitted by
Matthew Daubert

Problem E
A particle is moving with the given data. Find a formula for the position, s(t), of the particle at time t if a(t)=cos(t)+sin(t), s(0)=0, and v(0)=5.

Solution

We know that (integral)a(t) = v(t) and (integral)v(t) = s(t)

*    a(t) = cos(t) + sin(t)
(integral)a(t) = v(t) = sin(t) - cos(t) + C
v(0) = 5 = 0 - 1 + C so C = 6

**   v(t) = sin(t) - cos(t) + 6
(integral)v(t) = s(t) = -cos(t) - sin(t) + 6t + C
s(0) = 0 = -1 - 0 + 0 + C so C = 1

***  s(t) = -cos(t) - sin(t) + 6t + 1

Submitted by
Steven Emerson

Problem P
Compute the integral from 1 to 2 of (3x^2- {4\over x^2}) using calculus.

Solution

To compute the integral take the antiderivative of the function.
2                        2
S     3x^2 - (4/x^2) dx   =   I    x^3 + (4/x)
1                            1
Plug the two bounds in for x and subtract the top from the
bottom to find the value of the integral.
This =    8 + 2 - 1 - 4 = 10 - 5 =    5

Submitted by
Michael Pandolfo

Problem Q
For the curve y^3+5xy^2+x^4=-16 find
a) dy/dx
b) An equation of the line tangent to the curve at (-2,2)

Solution

a) dy/dx
(dy/dx both sides)   y^3+5xy^2+x^4=-16

3y^2dy/dx+5y^2+10xydy/dx+4x^3=0

(solve for dy/dx)   dy/dx(3y^2+10xy)= -5y^2-4x^3

dy/dx= (-5y^2-4x^3)/(3y^2+10xy)

b) An equation of the line tangent to the curve at (-2,2)
x=-2 and y=2.  So:  y=mx+b
where m is the slope which we already have in form of dy/dx

then dy/dx = (-5y^2-4x^3)/(3y^2+10xy) = m

m= [-5(2)^2-4(-2)^3]/[3(2)^2+10(-2)(2)] = -3/7

then in the equation y=mx+b we will solve for the value of b

2=-(3/7)(-2)+b so 2-(6/7)=b and b=(8/7).

Then the line tangent to the curve at (-2,2) is y=-(3/7)x+(8/7)

b:   y=-(3/7)x+(8/7)

Submitted by
Mehreen Qureshi
Comment from the management
We can ask Maple to draw a picture of the situation with the command
implicitplot({y^3+5*x*y^2+x^4=-16,y=-(3/7)*x+(8/7),x=-4..0,y=0..4);
and we should get something like the picture displayed, which certainly appears that the line found by Ms. Qureshi is tangent!

Problem S2
Compute the following limits. Give exact answers, as necessary in terms of well-known constants such as Pi, e, ln(3), sqrt{2}, etc.
a) limx-->0}[tan(17x)]/[sin(13x)]
b) limx-->infinity[sqrt(4+5x^4)]/[3+6x+7x^2]
c) limx-->1[ln(2x^2-1)]/[(x-1)ex]

Solution

a)lim tan(17x)/sin(13x)= L'H
x-0
0/0
lim [sec^2(17x)(17)] / [cos(13x)(13)] = 17/13
x-0

b) lim [(4+5x^4)^(1/2)]/[3+6x+7x^2] =
x-00 (infinity)
00/00   divide everything by x^2

lim [{(4+5x^4)(1/x^4)}^(1/2)]/[(3+6x+7x^2)(1/x^2)] = [(5)^(1/2)]/7
x-00


Submitted by
Patricia Sandoval