Wondering what everyone thinks about the common core math. I read that it's not necessarily meant to teach students to figure out an answer to a problem quicker, but it's more to teach children how to think. The rationale behind it is that kids don't need to know to do the calculations, per say, since a calculator can do all the work for them. The old traditional methods of adding, subtracting, multiplying and division are quicker, but many believe that kids just memorize the steps, the process by which to get the answer without even knowing why different things are performed. For example, why are numbers carried and borrowed?   As a long-time math tutor, I prefer the old traditional methods. I'd much rather use them and not allow kids to use calculators on the more simple math problems. Anyone have an opinion on the common core math methods?

When calculating the area under a curve over a certain interval, you can approximate it by getting the area of rectangles under the curve or equal widths. The more narrow the rectangle, the closer the approximation to the area.

The exact area can be found by integrating the function, which is the opposite of differentiation.

Suppose we want the area under the curve defined by the function f(x) = x^2 + 4x + 6 from [1, 4].

We integrate the function first. To do so we take 1/(exponent + 1)(coefficient and variable)^(exponent + 1). 

So x^2 becomes (1/3)x^3

4x becomes 2x^2

and 6 becomes 6x

Put it all together we have (1/3)x^3 + 2x^2 + 6x.  Now we substitute the values in the interval for x, first 4.

(1/3)(4^3) + 2(4)^2 + 6(4) = 77 1/3

Now substitute 1 for x to get

(1/3) + 4 + 6 = 10 1/3

Subtract 10 1/3 from 77 1/3 to get the final answer of 67.

Suppose the position of an object at time t is noted by the function s(t) = -4.9t^2 + 300.  The object is dropped from the top of a building. What is the average velocity of the object in the first three seconds after it is dropped?

The velocity, v(t), is found by taking the derivative, which is the rate of change of the object.

v(t) = -9.8t

Now we can used the mean value theorem, which is [f(b)- f(b)]/(b - a) for some interval a to b in which the function is differentiable. The interval is 0 to 3, so a = 0 and b = 3

f(3) = 255.9
f(0) = 300

(255.9 - 300)/(3 - 0) = -14.7 which is the average velocity. How can velocity be negative, one might think? The object is going downward, which makes it negative. Speed can only be positive but velocity can be both positive or negative depending on the direction.

Suppose f(x) = x^2 - 3x + 7, which is a parabola, and want to know the maximum and minimum on the interval [0,2]. 

First, we find the critical values, which is where the slope along the curve equals 0. To get that we take the derivative, set equal to 0 and solve for x.

f'(x) (derivative) = 2x - 3.

Set equal to 0 and we see that x = 3/2 or 1.5

Now, find the y coordinate of the endpoints of the interval and of the critical point.

f(0) = 7
f(2) = 5
f(1.5) = 4.75

Therefore the maximum on the interval is at 0,7 and the minimum on the interval is (1.5, 4.75).  In fact, since this is a parabola, the vertex is (1.5, 4.75), which is the minimum value of the curve.

suppose a car has a wheel with a radius of 15 inches. How many revolutions must the wheel make to cover a distance of 10 miles?

First you have to figure out what distance a wheel will cover in one revolution. That means we calculate the circumference of the wheel.

C= 2Pi(Radius)

C= 2(3.14)(15)

Therefore, the circumference is approximately 94.2 inches, which is 7.85 feet.

There are 52,800 feet in 10 miles. Therefore, there are 52800/7.85 = 6726 revolutions.

Suppose you have a conical tank that is 24 feet deep and 10 feet across. Water flows in at the rate of 10 cubic feet per minute. How fast is the depth rising when the depth of the tank is 8 feet?

We can use calculus to solve this problem.

The volume of a cone is 1/3(Pi)(Radius squared)(Height)

To solve this, use implicit differentiation. First we need to substitute a value for r in terms of h.  The we know that 5/12 = r/h, therefore r = (5/12)h.

V = (1/3)Pi(5/12 h)^2(h)
    = (1/3)Pi(25/144)h^3

Differentiate to get

V' = (1/3)Pi(75/144)h^2(h')

We know V' = 10, h = 8. Substitute those values into the equation and solve for h'.

10 =  (1/3)Pi(75/144)(8)h'

10 = 4.36h'

h' = 2.29

Therefore, the depth of the tank is rising at the rate of 2.29 cubic feet per minute.