Constructing Antiderivatives and areas

In summary, the first problem involves finding the ratio of areas of two parts of a square divided by the curve \sqrt{x} + \sqrt{y} = \sqrt{a}. The second problem involves finding the total cost of drilling a well x meters deep, given fixed costs of 1,000,000 riyals and marginal costs of 4000 + 10x riyals/meter.
  • #1
Jacobpm64
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0

Homework Statement


The origin and the point (a, a) are at opposite corners of a square. Calculate the ratio of the areas of the two parts into which the curve [tex] \sqrt{x} + \sqrt{y} = \sqrt{a} [/tex] divides the square.

Homework Equations


I'm sure there will be some use of A = bh. Perhaps maybe the pythagorean theorem if the square is cut exactly in half? I'm sure I'll be using a definite integral to find the areas.

The Attempt at a Solution


I have no clue where to go with this.

Homework Statement


In drilling an oil well, the total cost, C, consists of fixed costs (independent of the depth of the well) and marginal costs, which depend on depth; drilling becomes more expensive, per meter, deeper into the earth. Suppose the fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and the marginal costs are C'(x) = 4000 + 10x riyals/meter, where x is the depth in meters. Find the total cost of drilling a well x meters deep.

Homework Equations


I'm using I'll use a definite integral.

The Attempt at a Solution


Here's what I tried.
[tex] 1,000,000 + \int (from 0 to x) of 4000 + 10x [/tex]
Evaluating that:
[tex] 1,000,000 + [4000 + 10x - (4000 + 10(0) ) ] [/tex]
[tex] 1,000,000 + [4000 + 10x - 4000] [/tex]
[tex] 1,000,000 + 10x [/tex] riyals is the total cost.

I need some direction on the first one.. And I would like confirmation on the second one.
 
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  • #2
For the first one, I would suggest drawing your two areas out.

For the second one, is a fixed cost supposed to mean a limit? In other words, does fixed cost mean that C(x)<=1,000,000? If, on the other hand, it means that there is some cost that automatically comes in when drilling even a millionth of a unit down then you did it right.
 

1. What is the purpose of constructing antiderivatives and areas?

Constructing antiderivatives and areas is an essential part of calculus. It allows us to find the original function from its derivative and also helps in calculating the area under a curve, which is important in many real-world applications.

2. How do you construct an antiderivative?

An antiderivative of a function f(x) is a function F(x) whose derivative is equal to f(x). To construct an antiderivative, we use the reverse of the rules for finding derivatives. For example, if the derivative of a function is 3x^2, the antiderivative would be x^3 + C, where C is a constant.

3. What is the relationship between antiderivatives and definite integrals?

The definite integral of a function f(x) from a to b is the area under the curve of the function between a and b. This area can also be calculated by finding the antiderivative of f(x) and evaluating it at b and a, then taking the difference between the two values. Therefore, antiderivatives and definite integrals are closely related.

4. How do you find the area under a curve using antiderivatives?

To find the area under a curve using antiderivatives, we first find the antiderivative of the function. Then, we evaluate the antiderivative at the upper and lower limits of the area, and take the difference between the two values. This will give us the exact area under the curve.

5. Are there any practical applications of constructing antiderivatives and areas?

Yes, there are many practical applications of constructing antiderivatives and areas. In engineering, antiderivatives are used to find displacement, velocity, and acceleration of objects. In physics, they are used to calculate work, energy, and power. In economics, they are used to find marginal cost and marginal revenue. Overall, constructing antiderivatives and areas is an important tool in many fields of science and mathematics.

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