Problem 7 - Calculus

Question

Find where the curves in \(1-12\) intersect, draw rough graphs, and compute the area between them. $$ y=x^{2} \text { and } y=-x^{2}+18 x $$

Step-by-Step Solution

Answer

The curves intersect at (0,0) and (9,81), and the area between them is 243 square units.

Step 1: Find Intersection Points

To find the intersection points of the curves, set the equations for the two curves equal to each other: \ \[ x^2 = -x^2 + 18x \] \ Rearrange the equation to form: \ \[ x^2 + x^2 - 18x = 0 \] \ This simplifies to: \ \[ 2x^2 - 18x = 0 \] \Factor the quadratic equation: \\[ 2x(x - 9) = 0 \] \Setting each factor to zero gives the solutions: \\[ x = 0 \] and \[ x = 9 \]. \ The intersection points are at \(x = 0\) and \(x = 9\).

Step 2: Calculate y-values at the Intersections

Now, calculate the y-values at these intersection points using either of the original equations. Using \(y = x^2\): \ For \(x = 0\), \(y = 0^2 = 0\). \ For \(x = 9\), \(y = 9^2 = 81\). \ Thus, the intersection points are \((0,0)\) and \((9,81)\).

Step 3: Determine Bounds for Integration

The area between the curves is from \(x = 0\) to \(x = 9\).

Step 4: Set Up the Integral for Area Calculation

The area between two curves \(y = f(x)\) and \(y = g(x)\), from \(x = a\) to \(x = b\), is given by: \ \[ A = \int_{a}^{b} |f(x) - g(x)| \, dx \] \ For our problem, \(f(x) = -x^2 + 18x\) (the upper curve), and \(g(x) = x^2\) (the lower curve). The area \(A\) is: \ \[ A = \int_0^9 ( (-x^2 + 18x) - x^2 ) \, dx \] \ Simplify the integrand: \ \[ A = \int_0^9 (-2x^2 + 18x) \, dx \]

Step 5: Evaluate the Integral

To find the area, evaluate the integral: \ \[ A = \int_0^9 (-2x^2 + 18x) \, dx \] \ This can be split into two parts: \ \[ \int_0^9 -2x^2 \, dx + \int_0^9 18x \, dx \] \ Evaluate each part separately: \ \( \int -2x^2 \, dx = \frac{-2}{3}x^3 \) \ \( \int 18x \, dx = 9x^2 \) \ So, \ \[ \left[ \frac{-2}{3}x^3 + 9x^2 \right]_0^9 = \left[ \frac{-2}{3}(9)^3 + 9(9)^2 \right] - \left[ \frac{-2}{3}(0)^3 + 9(0)^2 \right] \] \ Calculate: \ \[ \left[ \frac{-2}{3}(729) + 729 \right] - 0 = \left[ -486 + 729 \right] \] \ \[ 243 \] \ Thus, the area between the curves is 243 square units.

Key Concepts

Intersection Points

Understanding intersection points is crucial when you want to find the area between two curves. Intersection points occur where the graphs of the equations cross each other. To find these points, set the equations for the two curves equal to each other. For example, with the curves given by:

• \( y = x^2 \)
• \( y = -x^2 + 18x \)

Set these equal to find the intersection:\[ x^2 = -x^2 + 18x \] By rearranging and simplifying, we solve: \[ 2x^2 - 18x = 0 \] Factor out common terms:\[ 2x(x - 9) = 0 \] This implies that \( x = 0 \) and \( x = 9 \). To find the y-values of these points, substitute these x-values back into either original equation, which gives you the points (0,0) and (9,81). These are the points where the two curves intersect each other.

Integral Calculus

Integral calculus is the key tool to find areas beneath and between curves. The idea is to sum up infinite small slices under a curve between two bounds. To find the area between two curves, subtract the lower curve from the upper one, and integrate across the desired interval. For curves described by \( y=f(x) \) and \( y=g(x) \), the area \( A \) from \( x=a \) to \( x=b \) can be calculated using:\[ A = \int_{a}^{b} [f(x) - g(x)] \, dx \]In this exercise, the area between \( y = -x^2 + 18x \) and \( y = x^2 \) from \( x = 0 \) to \( x = 9 \) is:\[ A = \int_0^9 ( (-x^2 + 18x) - x^2 ) \, dx = \int_0^9 (-2x^2 + 18x) \, dx \] Breaking this integral into two parts makes it simpler to solve, allowing us to use basic integration techniques on each term.

Quadratic Functions

Quadratic functions are a type of polynomial function with the highest degree of 2. They are usually expressed in the form \( ax^2 + bx + c \). The graphs of quadratic functions are parabolas that either open upwards or downwards.In this exercise:

• \( y = x^2 \) is a simple upward-opening parabola.
• \( y = -x^2 + 18x \) is a downward-opening parabola, modified by the linear term \( 18x \). This gives it a vertex pushed up compared to the standard parabola.

When analyzing or graphing these functions, identifying features such as the vertex (turning point) and direction of the curve helps understand their behavior and the regions they bound. As seen, these concepts apply not only to plot these curves but also to compute areas between them when they intersect.