Final Exam: Calculus & Modeling 1 (22M:021)   17 Dec 01	
Mike Winders & Keith Stroyan

Name: ________________  leave scores blank:
(({{1}, {   }}),({{2  }, {   }}),({{3  }, {   }}),({{4  }, {   }}),({{5 }, {   }}),({{  6   }, { }}),({{  7   }, { }}),({{ total }, { }}))

1 Differentiation Skills: Compute the following derivatives

a)  y = 5 (x^3) - (x^(1/3)),  find  (dy/dx)

b)  y  =  (e^x) Cos[x],  find  (dy/dx)

c)  y = Log[(x^2) + 2 x + 1],  find  (dy/dx)

d)  f[x]  =  (x^2) (e^(3  x)),  find  f'[x]

2 Integration Skills: Compute the following:

a)   \(\(\(\ \)\(\(∫\((5\ x\^3 - \@x\%3)\) dx\)\(\ \)\(=\)\)\)\)

b)  (∫_(π/6)^(π/4))Cos[2 θ] dθ =

c)  (∫_0^1) ((x^2)/(1 + 
                      (x^3))) dx=

d)  (∫_1^e)x Log[x] dx

e)  ∫ Sin[θ] Cos[θ]dθ =

3 Knee Jerk

A function of time,  y =  y[t]  satisfies:

y[0]  = - 3
 &				
	(dy/dt) = 2  y

Give the exact value of  y[4] = ? and explain your formula.  (Note:  Log[2]  is exact, 0.693147, is approxiamte.  Sin[1]  is exact, 0.841471,  is approximate.)

4  Tangent Lines

The explicit curve  y = f[x]  is shown below with  f[x] = 5  (x^3) - 3  (x^5).  Calculate the x-values where the graph has a horizontal tangent line and sketch those tangents on the graph.

f[x_] = 5  (x^3) - 3  (x^5);
Plot[f[x], {x, -3/2, 3/2}, 
        AxesLabel -> {"x", "y"}];

5 Shapes and Derivatives

The graph of  y =  ((x (((x - 2))^3) (x-6))^2)= -288 x + 528 (x^2) - 368 (x^3) + 120 (x^4) - 18 (x^5) + 
        (x^6)  is shown below.  The x-values indicated at the bottom are the ONLY places where either  (dy/dx)= 0 or  (((d^2) y)/(dx^2)) = 0.   (x = 0.44, 0.88, 2.0, 3.6, 5.5, 6.0)  Make a complete shape table for this function showing where y  is increasing and decreasing and where  y = f[x]  is convex up or down.

Clear[x, y];
y[x_] = x (((x - 6))^2) (((x - 2))^3)
Expand[y[x]]
NSolve[(∂_x) y[x] == 0]
\(NSolve[∂\_{x, 2}y[x] == 0, x]\)
curve = 
        Plot[y[x], {x, - .2, 6.5}, 
          AxesLabel -> {"x", "y"}];

Show[curve, 
        Graphics[{Line[{{ .438, y[ .438]}, { .438, -55}}], 
            Line[{{ .88, y[ .88]}, { .88, -55}}], 
            Line[{{2, y[2]}, {2, -55}}], 
            Line[{{3.65, y[3.65]}, {3.65, -55}}], 
            Line[{{5.47, y[5.47]}, {5.47, -55}}], 
            Line[{{4.56, y[4.56]}, {4.56, -55}}], 
            Line[{{6, y[6]}, {6, -55}}]}], 
        Graphics[{Text[".44", { .438, -65}], 
            Text[".88", { .88, -65}], Text[2. , {2, -65}], 
            Text[3.6, {3.65, -65}], Text[4.6, {4.56, -65}], 
            Text[5.5, {5.47, -65}], 
            Text[6.0, {6, -65}]}]];

6 Max-min

Find the max and min of the function  f[x] = (x^3) - 3 x + 4  for  x  in the interval  [-2,3].  Does  f[x]  have both a max and min?  Explain your reasoning carefully.

f[x_] = (x^3) - 3 x + 4;
f'[x]
Solve[f'[x] == 0]
Plot[f[x], {x, -2, 3}]

f[-1]
f[-2]
f[3]
f[1]

7 A Differential Equation describing something

Fact 1:  A certain chemical reaction proceeds at a rate proportional to the concentration of a compound  C.  The reaction is faster when the concentration of compound  C  is higher and slower as the concentration of  C  is lower and the relation between reaction rate and concentration is linear for this particular compound.

Fact 2:  This reaction uses the compound  C  up, so the faster the reaction is going, the more compound  C  is being used per second.

Fact 3:  The reaction begins with a concentration of  2.0 millimolars at time  0.

Fact 4:  3 seconds later, the concentation is  1.5 millimolars.

Let
	c = the concentration of compound  C  in millimolars
	t = time in seconds beginning with  t=0 at the start of the reaction

What is the concentration  c  at  time  t=9 sec?

HINT a)  Use Facts 1 & 2 to describe the reaction rate in terms of the derivative  (dc/dt)

(dc/dt)  =  ??

HINT b)  Use Fact 3  and  part  (a)  to find a general formula c = c[t] = ???  valid for all   t ≥0.

c[t_] =  2 E^(-k t)
\(		Solve[c[3] == 3/2, k]\)

HINT c)  Use Fact 4 to find a specific formula c[t]=??