Calculus

'20-'21 Calculus Students:
here's your supply list for class.
I'm looking forward to August! Enjoy the remainder of your summer.

  • binder with loose leaf paper OR spiral notebook with pockets and enough paper to last all school year-for MATH only! (A 1-inch binder with a zippered pencil pouch works best.)
  • pencils, pens (dark blue or black for regular classroom purposes),
  • large eraser
  • colored pencils or an assortment of colored pens for graphing
  • graph paper
  • protractor
  • box of facial tissues and container of disinfecting wipes
Week Six Assignments (Friday, May 15-Thursday, May 21)

Web-Based

Khan Academy Calculus
Sign up using Class Code YCP5DUE6
I will post assignments relating to Mean Value Theorem.

Non Web-Based
For each equation, find all relative extrema.

1. f(x) = (x-5)²

2. g(x) = x³-3x²+3

3. h(x) = x⁴-4x³+2

Solutions to Week Six

To find the relative extrema, first use the derivative to find the critical numbers.
Then, use the second derivative to determine if the critical points represent a relative minimum (if f"(x)>0), relative maximum (if f"(x)<0), or neither (if f"(x)=0).

1. f'(x)= 2(x-5)                      f"(x)= 2 ⇒ f"(5)=2, which is >0.      f(5)=0
Critical Number: x=5.                                              The point (5,0) is a relative minimum.

2.  g'(x)= 3x²-6x  ⇒ 3x(x-2)=0    Critical Numbers x=0 or x=2.        g(0)=3 and g(2)= -1
      g"(x)= 6x-6  ⇒ g"(0) = -6     g"(2) = 6    
          The point (0,3) is a relative maximum.  The point (2,-1) is a relative minimum.

3. h'(x)= 4x³-12x²    ⇒ 4x²(x-3)=0   Critical numbers x=0 and x=3.  h(0)=2 and h(3)=-25/
      h"(x)=12x²-24x ⇒   h"(0)=0    h"(3)=36
                            The point (0,2) is not an extremum. The point (3, -25) is a relative minimum.


Week Five Assignments (Friday, May 8-Thursday, May 14)

Web-Based

Khan Academy Calculus
Sign up using Class Code YCP5DUE6
I will post assignments relating to Mean Value Theorem.

Non Web-Based
Related Rates
Hint: the formula for the volume of a cone is V= (1/3)𝝿r²h.
Related Rates
1. Water flows into a conical tank at the constant rate of 3 cubic meters per second. The radius of the cone is 5 meters and its height is 4 meters. Let h(t) represent the height of the water above the bottom of the cone at time t. Note: the formula for the volume of a cone is V= (1/3)𝝿r²h.

a) Find dh/dt (the rate at which the water is rising in the tank) when the tank is filled at a height of  2 meters, and

Solution: dh/dt = 12/(25𝝿) meters per second.

b) d²h/dt², at the same height of 2 meters.

Solution:  d²h/dt² = -144/625𝝿² meters per second per second
Hint: the formula for the volume of a cone is V= (1/3)𝝿r²h.



Week Four Assignments (Friday, May 1-Thursday, May 7)

Web-Based

Khan Academy Calculus
Sign up using Class Code YCP5DUE6
I will post assignments relating to Mean Value Theorem.

Non Web-Based
Related Rates
1. All edges of a cube are expanding at a rate of 6 centimeters per second. How fast is the volume changing when each edge is a) 2 centimenters, and b) 10 centimenters.
Solutions: V=s³      dV/dt = 3s²

a) s=2cm   s'=6 cm/sec.            dV/dt= 3(2²)(6) = 72cm³/sec

b) s=10 cm.         dV/dt= 3(10²)6 = 1800 cm³/sec

2. The conditions are the same as in the previous problem. Determine how fast the surface area is changing when each edge is a) 2 centimeters,  and b) 10 centimeters.
Solutions: SA= 6s²        dA/dt= 12ss'

a) s= 2 cm   s'= 6 cm/sec     dA/dt= 12(2)6 = 144 cm²/sec

b)  dA/dt= 12(10)6 = 720 cm²/sec



Week Three Assignments (Friday, April 24-Thursday, April 30)

Web-Based

Khan Academy Calculus
Sign up using Class Code YCP5DUE6
I will post assignments relating to Mean Value Theorem.

Non Web-Based
For each function, f(x),
a. find the critical numbers of f, if they exist, 
b) find the open intervals on which the function is increasing or decreasing, and
c) locate all relative extrema.

1. f(x)= x³-6x²+15,

2. f(x) = x/(x+3)

3. f(x) = (x+4)/x²

Solutions to Week Three

1a) f'(x) = 3x²-12x       Critical Numbers: x=0 and x=4

b) Increasing: (-∞,0) and (4,∞); Decreasing (0,4)

c) Relative Maximum (0, 15); Relative Minimum (4,-17)

2a) f'(x)= 3/(x+3)²        Critical Number: x=-3 is undefined, therefore, there are no critical numbers.

b) Increasing (-∞,-3) and (-3,∞) Discontinuity at -3.

c) No relative maximum or minimum.

3a: f'(x)= (-x-8)/x³       x=0 is undefined.
Critical number: x=-8

b) Decreasing: (-∞,-8) and (0,∞)       Increasing: (-8,0)

c) Relative Minimum: (-8, -1/16)

Week Two Assignments (Friday, April 17-Thursday, April 23)

Web-Based

Khan Academy Calcul
Sign up using Class Code YCP5DUE6
I will post assignments relating to Mean Value Theorem.

Non Web-Based

Find the first and second derivatives of

  1. 1. y=5x² – 18

  1. 2. y=5x⁵-18x+12

  1. 3. f(x)= sin 3x


Solutions to Week 2

1. y'=10x;  y"= 10

2. y'= 25x⁴-18;  y"= 100x³

3. f'(x)= 3 cos(3x);  f"(x)= -9 sin(3x)
Week One Assignments (Friday, April 10-Thursday, April 16)

Web-Based

 Khan Academy Calculus
Sign up using Class Code YCP5DUE6
I will post assignments relating to Mean Value Theorem.

Non Web-Based

Find the values that satisfy Rolle’s Theorem:  

  1. y=x³ −x² −4x+3; [−2,2]
  2. y = sin (2x); [−𝝿, 𝝿]
Solutions to Week One
To apply Rolle's Theorem, test the endpoints of the closed interval for equality.
I'll test these points and take it one step further by identifying the turning points of each function. In other words, I'll find the points at which the derivative is equal to zero. I'm going to convert y into f(x).

1. Since f(-2)=-1 and  f(2)=-1, Rolle's Theorem applies.
To find the turning points of the function: 
f'(x)= 3x²-2x-4
Using the Quadratic Equation, y= [-b±√(b²-4ac)]/2a, the zeros of the derivative simplify to  x=[1±√13]/3.

2. Since f(-𝝿)=0 and f(𝛑)=0, Rolle's Theorem applies.
f'(x)=2cos(2x)
2cos(2x)=0 at  x=±3𝝿/4 and ± 𝛑/4.

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