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)

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.

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

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

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)

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

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

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) lim_x-->0}[tan(17x)]/[sin(13x)]
b) lim_x-->infinity[sqrt(4+5x^4)]/[3+6x+7x^2]
c) lim_x-->1[ln(2x^2-1)]/[(x-1)e^x]

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