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)