How Do You Graph and Calculate the Area of a Region Defined by Two Inequalities?

  • Thread starter Thread starter TheShapeOfTime
  • Start date Start date
  • Tags Tags
    Graphing
AI Thread Summary
To graph the region defined by the inequalities x^2 + y^2 - 2x + 4y - 5 <= 0 and x + y - 1 >= 0, first rewrite the first inequality as (x-1)^2 + (y+2)^2 <= 10, representing a circle centered at (1, -2) with a radius of √10. The second inequality can be expressed as y >= -x + 1, indicating the area above the line y = -x + 1. The solution region is the overlap of the shaded areas from both inequalities. To calculate the area, find the area of the circle and subtract the area of the triangle formed by the line and the axes. This process will help in accurately determining the area of the defined region.
TheShapeOfTime
"Graph the region defined by the inequalities x^2 + y^2 - 2x + 4y -5 &lt;= 0 and x + y - 1 &gt;= 0. Determine the area of the region defined bythe graph."

I'm confused about the graphing part, I have no idea where to begin.
 
Physics news on Phys.org
Graph the region defined by each separate inequality. The area of overlap between the two areas is the region the question is asking about.
 
Can you provide a step-by-step guide or some tips on how to graph these equations?

Sure, here is a step-by-step guide on how to graph these two equations and determine the area of the region.

Step 1: Rewrite the inequalities in slope-intercept form. This will make it easier to graph the equations. The first inequality, x^2 + y^2 - 2x + 4y -5 <= 0, can be rewritten as (x-1)^2 + (y+2)^2 <= 10. The second inequality, x + y - 1 >= 0, can be rewritten as y >= -x + 1.

Step 2: Plot the center point of the first inequality, which is (1,-2). This is where the two equations intersect.

Step 3: Plot the radius of the first inequality, which is √10. This is the distance from the center point to the edge of the circle.

Step 4: Shade in the region inside the circle, including the boundary line.

Step 5: Plot the line y = -x + 1, which is the boundary line of the second inequality.

Step 6: Shade in the region above the line, including the boundary line.

Step 7: The shaded region where the two inequalities overlap is the solution to the system of inequalities. This is the region that satisfies both equations.

Step 8: To determine the area of the region, you can use the formula for the area of a circle, A = πr^2, to find the area of the circle inside the shaded region. Then, use the formula for the area of a triangle, A = 1/2bh, to find the area of the triangle formed by the boundary line and the x and y axes. Finally, subtract the area of the triangle from the area of the circle to find the total area of the shaded region.

I hope this helps guide you through the process of graphing and finding the area of the region defined by these two equations. Remember, practice makes perfect, so keep practicing graphing and solving systems of inequalities to improve your skills.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top