Semester Recap

     I'm really surprised by how well I've done in calculus. I've always had a hard time with math, so I really think that my success this semester had to with the style of teaching. I think my strongest area in calculus is derivatives, but sometimes the chain rule confuses me. My weakest area in calculus is definitely anything that has to do with finding the domain and range of a function. I'm also pretty bad at solving limits algebraically.
     My major is biology and I'm also minoring in biochemistry. Calculus is required for my major because we are required to take physics, which deals with a lot of calculus. Calculus is also present in chemistry which pretty much goes hand in hand with biology. Biology uses calculus to interpret graphs of data, measure pH levels and calculate growth and decay rates. Rates of change are also important in biology. Applied optimization is also an important part of calculus that is used frequently in biology.  Calculus is used in chemistry for all of the same things including calculating half-lifes of different substances. I'm not really sure what other aspects of calculus could be used in my major that are not used already. It seems as though I can find an example of pretty much every topic we have covered somewhere in biology, but I'm sure that there are topics in calculus 2 that could be used somewhere in biology.


Problems:

1. Find the limit as x approaches 0 of x+tan(x) / sin(x).
Since this limit gives you 0 over 0, you have to use L'Hopital's rule. By taking the derivative of the numerator and denominator, the function becomes: 1+sec^2(x) / cos(x). Now you can plug 0 in for x, which finds that the limit is equal to 2.

2. Find the limit of (2x^2+x-10) / (2x^5-40x+16) as x approaches 2.
(2(2)^2 + 2 - 10)/(2(2)^5 - 40(2) + 16) = 0/0.
By taking the derivative: limit as x approaches 2 of (4x+1) / (10x^4-40)
Plug in 2: (4(2) + 1) / (10(2)^4 - 40) = 9/120 = 3/40

3. Find the integral:
 ∫x^5+3x+2 dx = (1/6)x^6 + (3/2)x^2 + 2x + k

4. Find the integral evaluated from 1 to 3.
  ∫ (10/x)+4x^3 dx = 10*ln|x| + x^4 + k
10*ln(3) + 3^4 + k - (10*ln(1) + 1^4 + k) = 92.98

5. Find the integral evaluated from 0 to 2.
∫ (x^2  + cos(x) - 4x = (1/3)x^3 + sin(x) - 2x^2 + k
1/3(2)^3 - 2(2)^2 + k - (1/3(0)^3 + sin(0) - 2(0)2 + k) = -4.424

6. Find the area between the two curves:
f(x)= 3x^2
g(x) = 4-x.

The points where the functions intersect are: (-1,5) and (1,3)

∫3x^2 = x^3 + k
(1)^3 - ((-1)^3)) = 2

∫4-x = 4x - (1/2)x^2 + k
4(1)-((1/2)(1)^2) - (4(-1) - (1/2)(-1)^2) = 8

Area between the curves: 8 - 2 = 6

7. Find the area between the curves evaluated from -2 to 0.
f(x) = 7cos(x)
g(x) = 3x^2 + 7x

∫7cos(x) = 7 sin(x) + k
7*sin(0) - 7*sin(-2) = 6.36

∫3x^2 + 7x = x^3 + (7/2)x^2 + k
3(0)^2 + 7(0) - (3(-2)^2 + 7(-2)) = 2

The area between the curves = 6.36 - 2 = 4.36

Roller Coaster -

F(x)= x^6 - 4x^5 + 3x^4 + 6x^3 - 4x^2 - 12x

F'(x)=6x^5 - 20x^4 + 12x^3 + 18x^2 - 8x - 12=0

x= 1.15, -0.88, 1, 0.816, -1.5

f(1.15) = 1.15^6 - 4(1.15)^5 + 3(1.15)^4 + 6(1.148)^3 - 4(1.15)^2 - 12(1.15)

          = -10.43

f(-0.88)= (-0.88)^6 - 4(-0.88)^5 + 3(-0.88)^4 + 6(-0.88)^3 - 4(-0.88)^2 - 12(-0.88)

           =-13.57

f(1) = 1^6 - 4(1)^5 +3(1)^4 + 6(1)^3 - 4(1)^2 -12(1)

      = -10

f(-0.816) = -0.816^6 - 4(-0.816)^5 + 3(-0.816)^4 + 6(-0.816)^3 - 4(-0.816)^2 - 12(-0.816)

           = -3.73

f(-1.5) = (-1.5)^6 - 4(-1.5)^5 + 3(-1.5)^4 + 6(-1.5)^3 - 4(-1.5)^2 -12(-1.5)

          = 46.82

The critical points for this function, which will have 3 local minima and 2 local maxima are: (1.15,-10.43), (-0.88,-13.57), (1,-10), (-0.816,-3.73) & (-1.5, 46.82)

4.6: Applied Optimization

Applied Optimization is pretty tough,
finishing the homework has been rough.
But this topic is prevalent in many fields,
so I should probably keep my dislike concealed.

Problem:

#9. Suppose 600 ft. of fencing are used to enclose a corral in the shape of a rectangle with a semi-circle whose diameter is a side of a rectangle. Find the dimension of the corral with the maximum area.

Perimenter=2x+y+(pi*y)/2
600=2x+y+(pi*y)/2
x=(600-y-(pi/2)*y)/2

A(x)=(xy+pi(y/2)^2)/2
      =(600y-(y^2)+pi/2*y^2)/2 + 2+(pi*y)/8
      =300y-(y^2)/2- (1/8)*pi*y^2

A'(x)= 300 - (2y/2)-(1/4)*pi*y
A'(x)= 300-y-(1/4)*pi*y =0

y=1200/(4+pi)
x=(600-y-(pi/2)*y)/2

  =600/(4+pi)

So in order to get the maximum amound of space,  the dimensions of the corral need to be (600/(4+pi), 1200/(4+pi))

4.5: Graph Sketching and Asymptotes

Graph sketching depends on points of transition;
points where the sign of the slope changes position.


Problem:

1. Perform all necessary steps needed to sketch the graph of the function:
f(x)=x^3-3x^2

f'(x)=3x^2-6x =0
3x(x-2)=0
3x=0; x=0 & (x-2)=0;x=2

f(0)=0^3-3(0)^2 = 0
f(2)=2^3-3(2)^2 = -4

Critical Points: (0,0) & (2,-4)

For the sign graph of the first derivative: f'(-1)=9 - indicating a positive slope, f'(1)=-3 - indicating a negative slope and f'(3)=9 - indicating a positive slope.


f''(x)= 6x-6 =0
x=1
f(1)=1^3-3(1)^2 = -2

Point of Inflection: (1,-2)

For the sign graph of the second derivative: f''(0)=-6 - indicating a negative slope, f''(2)=6 - indicating a positive slope.

When the sign graph of the first and second derivatives are combined, the new sign graph looks like this: +-  0  --  1  -+  2  ++. The two zeros of the graph are located at 0 and 2 and the point of inflection, where the slope of the graph begins to change direction is at 1. Using the combined sign graphs from the first and second derivatives, I was able to draw a sketch of the graph of this function:

4.4: The Shape of a Graph

The shape of the graph is either up or down,
the shape of a smile or the shape of a frown.
The graph will contain a few points of inflection,
it's really easy to do if you follow directions.
To find the shape use the second derivative test,
Using a sign chart will end your quest.


Problem:
1. Using the equation f(x)=x^4-4x^3, find the second derivative, the critical points at the second derivative and where the shape of the graph is concave up/down.

f(x)=x^4-4x^3

f'(x)=4x^3-12x^2

f''(x)=12x^2-24x

f''(x)=12x^2-24x=0
12x(x-2)=0
12x=0; x=0
x-2=0; x=2

f(0)=0^4-4(0)^3=0. The first critical point is (0,0)
f(2)= 2^4-4(2)^3=-16. The second critical point is (2,-16)

To determine the sign of the graph before and after each critical point, I picked the numbers -1, 1, and 3. At f''(-1) the slope of the line is positive, at f''(1), the slope of the line is negative and at f''(3) the slope of the line is positive. This indicates that the shape of the graph at -1 is concave down and at 1 it is concave up.

4.3: The Mean Value Theorem and Monotonicity

Use the derivative to test the sign,
which will help determine the graph's design.
The change in sign will determine a local max or local min,
finally figuring this out made me grin!

Problem:

1. Find the critical point(s) and apply the first derivative test. Is there a local max or a local min at the critical point(s)?

f(x)=x^2+6x-7
f'(x)=2x+6=0
f'(x)=2(x+3)=0
2 can't be equal to zero.
x+3=0, x=-3.

F(x)=(-3)^2+6(-3)-7
f(x)=9+(-18)-7=-16

Critical point: (-3,-16)

I picked points to the left and right of -3. I used -4 and -2. I plugged -4 in for x in the derivative of the original function, and got -2. I plugged -2 in for x in the derivative of the original function and got 2. This means that there is a local minimum at -3, because the sign changes from - to +.

4.2: Extreme Values

Finding the derivative is one of my favorite parts,
they're used to find extrema on various charts.
The critical points are easy to find,
they're the horizontal tangent line.


Problem:

1. Find the critical point(s) at f(x)=9x^2-5x+4.
f'(x)=18x-5
f'(x)=18x-5=0
f'(x)=18x=5
x=5/18.

f(x)=9(5/18)^2-5(5/18)+4 = 3.305
y=3.305

The critical point for this function is ((5/18),3.305)