1. Find an equation for the line which passes through the point (2,-3) and is perpendicular to the line that connects the points (1,1) and (-4,0).

   Solution: The line through (1,1) and (-4,0) has slope

   1 - 0 1 - (-4) = 1 5

   Therefore the perpendicular line has slope -5. The line of slope -5 through the point (2,-3) is

   y - (-3) x-2 = -5

   which simplifies to y+3 = -5(x-2) or y = 7 - 5x.

2. Find the slope of the curve y = f(x) where f(x) = x² + 5x at the point (x,x² + 5x) by computing

   lim h goes to 0  f(x+h) - f(x) h

   Solution: In this case f(x+h) = (x+h)² + 5(x+h), so

   slope = lim h goes to 0  (x+h)2 + 5(x+h) - x2 - 5x h = lim h goes to 0  x2 + 2xh + h2 + 5x + 5h - x2 - 5x h = lim h goes to 0  h(2x+h+5) h = lim h goes to 0  2x + h + 5 = 2x + 5

1. If f(x) = x²⁷ - 5x⁶² + 3x - 7, then find f'(x).

   Solution: f�¢(x) = 27x²⁶ - 5·62 x⁶¹ + 3 = 27x²⁶ - 310x⁶¹ + 3.

2. 1. Find the indefinite integral of x⁵ - 3x¹⁰ + 2.

      Solution: The indefinite integral is x⁶/6 - 3x¹¹/11 + 2x + c

   2. Find the antiderivative of 5x³ - 7x which has the value -1 when x = 1.

      Solution: The antiderivative of 5x³ - 7x is 5x⁴/4 - 7x²/2 + c. We must choose c so that

      5x4/4 - 7x2/2 + c = -1 when x = 1.

      Thus we must have

      5/4 - 7/2 + c = -1, which leads to c = 5/4.

      So the answer is 5x⁴/4 - 7x²/2 + 5/4.

3. A ball is thrown straight up from the top of a 72-foot building with an initial velocity of 128 ft/sec. When will the ball reach its maximum height and what will that height be?

   Solution: As shown in class, the position function for an object falling near the surface of the earth, propelled only by gravity, will be

   s(t) = -16t2 + v0t + s0,

   where s is measured in feet, t is measured in seconds, v₀ represents the initial velocity, and s₀ represents the initial position. (Here we also assume that the positive direction is away from the surface of the earth.)

   So in our case s(t) = -16t² + 128t + 72 and v(t) = s'(t) = -32t + 128. The ball reaches its maximum height at the instant when the velocity decreases to 0. This occurs when 128 = 32t, or t = 4 sec. Thus the maximum height is s(4) = -256 + 512 + 72 = 328 feet.

Quiz 3, September 15, 1998

1. For

                                           
                x                     
   f(x) = -----------       and      g(x) = sqrt(x^2 - 1)
            x^2 - 1                          
   
   

   then
   • The domain of f is all real numbers except 1 and -1 (since the only trouble arises when we divide by zero, when x^2 - 1= 0.)

   • The domain of g is (-infinity, -1] U [1, infinity) since we can only take square roots of nonnegative numbers and x^2 - 1>0 if and only if |x| > 1, that is if x is greater than or equal to 1 or less than or equal to -1.

   • f(2) = 2/(4-1) = 2/3.

   • g(3x) = sqrt((3x)^2 - 1) = sqrt(9x^2 - 1)

   • f composed with g at x is equal to

      f(sqrt(x^2-1)) =     sqrt(x^2 - 1)         sqrt(x^2 - 1) 
                      ---------------------- = -------------------
                       (sqrt(x^2 - 1))^2 - 1        x^2 - 2
     

2. Note that 2*Pi/3 is in the second quadrant, 5*Pi/6 is in the second quadrant, and 7*Pi/4 is in the fourth.
   • sin(2*Pi/3) = sqrt(3)/2

   • sec(Pi/6) = 1/cos(Pi/6) = 1/(sqrt(3)/2)= 2/sqrt(3) = 2sqrt(3)/3

   • tan(5*Pi/6) = sin(5*Pi/6)/cos(5*Pi/6) = (1/2) / (-sqrt(3)/2) = -1/sqrt(3) = -sqrt(3)/3

   • cos(7*Pi/4) = 1/sqrt(2) = sqrt(2)/2

3. The relationship between arclength and angle in a circle is given by the formula

   Arc Length = (Angle)(Radius), where the angle is measured in radians.
   

   Thus in this example, Arc Length = (Pi/8)(12) = 3*Pi/2 or about 4.5.

Quiz 4

Solutions distributed in class. Questions:

 The following functions are undefined at x = 2.  Could they
be defined at x = 2 to give a continuous function.  Explain.

f(x) = |x-2|        f(x) = 3x^2 - 12         f(x) = x^2 - x - 6
       -----               ---------               -------------
        x-2                  2 - x                     x - 2



 Use the definition of derivative to show that Dx(sqrt x) = 1/(2sqrt(x)).


 Find an equation for the tangent line to 
 
y =  5x + 4
    --------          at the point (1, 9/2).
    x^2 + 1




Quiz 5 

 Compute the limit of (2*theta)/sin(5*theta) as theta goes to 0.
       2t        2 * 5t         (2/5)*  (5t)
     -------  = ---------   =          ------
     sin(5t)     5*sin(5t)             sin(5t)

So taking the limit as t goes to zero, the (2/5) comes out and
the quotient (5t)/sin(5t) goes to 1.  Thus the limit is 2/5.



Compute dy/dx if 
(a) y = (5x^2 - 3x)^17

      dy/dx = 17(5x^2 - 3x)^16 * (10x - 3)       (Chain Rule)

(b) y = sin( (2x+1)/(3x - 5) ) 

      dy/dx = cos( (2x+1)/(3x-5) ) * (3x-5)(2) - (2x+1)(3)
                                     ---------------------
                                          (3x-5)^2

(c) y = sec(x^2sin(x))

       dy/dx = sec(x^2*sin(x)) tan(x^2*sin(x) ) * (x^2*cos(x) + 2x*sin(x))


If y = 3x^2 - 2x and x1 = 1 then find Delta y corresponding to
Delta x = 0.1.  What numerical value will (Delta y)/(Delta x) approach
as Delta x goes to 0?

x1 = 1        so  y1 = 3(1^2) - 2(1) = 1
x2 = 1.1      so  y2 = 3(1.1)^2 0 2(1.1) = 1.43

So Delta y = 0.43 when Delta x = 0.1.

The quotient (Delta y)/(Delta x) will approach the slope of
the curve at x1 = 1.  In this case the slope of the curve is
given by 6x - 2, so when x = 1 we get a slope of 4.



                                                     Quiz 7


Problem 1:  Use the tangent line approximation
to estimate 1222/3.  Hint:  1252/3 is easy to compute.
Compare your approximation to the actual value.


Solution:  We work with f(x) = x2/3 whose
derivative is Df(x) = (2/3)x-1/3.  Then





1222/3 = f(125 - 3)  is about  f(125) + Df(125)(-3) = 25 + 

2

 3


1

 5

(-3)  = 25 -  

4

 10

 = 24.6.



The actual value, computed with calculator, is 24.59838.


Problem 2:  A hemispherical bowl of radius 10in
is filled with water to a depth of x inches.  The volume V 
of water in the bowl (in cubic inches) is given by





V = 

Pi

 3

(30x2 - x3)



If the depth of water is measured to be 5�±1/16 inches, then
compute the volume together with an estimate for the error.


Solution:  First compute DV(x) = (Pi/3)(60x - 3x2)
or Pi(20x - x2).  Then





V(5�±1/16)  is about  V(5) �±DV(5)/16 = (Pi/3)(625) �±(75Pi/16) = 654.50 �±14.73





Problem 3:  Find the maximum and the minimum of
f(x) = 3x4 - 4x3 - 12x2 + 5 on the interval [-2,3].


Solution:  The derivative of f is





Df(x) = 12x3 - 12x2 - 24x = 12x(x2 - x - 2) = 12x(x-2)(x+1)



so the critical points are the endpoints -2 and 3 together
with the stationary points 0, 2, and -1.  Computing function
values gives







f(-2) = 37,  the maximum 

f(3) = 32, 

f(0) = 5, 

f(2) = -27,  the minimum 

f(-1) = 0