Two Area problems.

Exercise: Find the upside down parabola so that the area between it and the parabola is 100. Draw a diagram.

Solution.

Set up the functions which bound the region.

> restart;

> f := x -> a-x^2;

> g := x-> x^2;

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Find the points where the functions cross.

> sol := solve(f(x)=g(x),x);

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Get the area of the region. By inspection, we see that sol[2] is the left endpoint where the functions cross.

> area := int(a-2*x^2,x=sol[2]..sol[1]);

> sol;

Solve an equation.

> sol2 := fsolve(area=100,{a});

> assign(sol2);

> plot({f ,g }, sol[2]..sol[1]);

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Another area problem: The parabola is tangent to the graph of at two points and the area of the region bounded by their graphs is 10. Find a, b, and c. Make a sketch.

Solution :

The axis of the parabola is . That is also the axis of , so , or . The point where the slope of the parabola is 1 is on both graphs. Call the point [x0,y0]. Then and .

> restart;

> eq1 := x0-1 = a*x0^2 -6*a*x0 + c;

> eq2 := 1 = 2*a*x0 - 6*a;

> ac := solve({eq1,eq2},{a,c});

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Finally, the area between the curves is 100, so the righthand half is 50.

> eq3 := Int (a*x^2-6*a*x+c-(2+x-3),x=3..x0)=50;

> eq3 := int (a*x^2-6*a*x+c-(2+x-3),x=3..x0)=50;

> sol :=solve(subs(ac,eq3),x0);

> assign({x0=sol[1]});
assign(ac);

> plot({2+abs(x-3),a*x^2-6*a*x+c},x=sol[2]-2..sol[1]+2,color=black);

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