Preview A.11: Calculus and Geometry Math 1510 Objective: The properties of the p

Question

Preview A.11: Calculus and Geometry Math 1510 Objective: The properties of the parabola (and other conic sections) were known to Euclid and Apollonius over 2000 years ago. In this activity you will use calculus to rediscover an interesting property of the parabola. Background: This activity is based on problem 15 in the Problems Plus section on page 363. You will also need to recall the formula for the area of a trapezoid: A==( b, +b, Jh Steps: 10 T Given a line y = mx +b with positive y-intercept, the line will intersect the parabola y=x in two points. We want to find the inscribed triangle with maximal area. a) For the line y -2x +3, find the coordinates of the intersection points A and B. -4 b) Let P = (x, y) be a point on the parabola between the intersection points A and B. Find a formula for the area of the triangle A ABP as a function of x. Hint: The area of the triangle can be found by taking the area of a big trapezoid and subtracting the areas of two smaller trapezoids. See the second figure. AREA(x)= c) Use calculus to find the coordinates of the point P that maximizes the area. Page 1 of 2

Explanation / Answer

Part (a)

We have to find intersection points of line y=-2x+3 and parabola y=x2

for finding intersection we find the values of x by setting both the functions equal:

x2 =-2x+3

x2+2x-3 =0

(x+3)(x-1) = 0

We get x=-3 and x=1

Plugging these values in the parabola we get corresponding y's as:

For x=-3 we get y=9

For x=1 we get y=1

Therefore, the required intersection points are

A = (-3,9)

B = (1,1)

Part (b)

Area = AreaBig Trapezoid - (Areasmall trapezoid 1+Areasmall trapezoid 2)

Area = 1/2 (9+1)(4)-(1/2 (9+y)(3+x)+1/2 (1+y)(1-x))

Area = 20 - 1/2 (9+x2)(3+x)-1/2 (1+x2)(1-x)

A (x) = 20 -(1/2)(27+9x+3x2+x3+1-x+x2-x3)

A (x) = 20-(1/2)(28+8x+4x2)

A (x) = 6-4x-2x2

Part (c)

For finding maximum area we can use derivatives.

Extrema occurs at those values of x where derivative is zero.

So we set:

A'(x) = 0-4-4x=0

4x=-4

x=-1

Corresponding value of y is:

y=(-1)2=1

Hence area is maximum for point P=(-1,1)

So maximum area = 6-4 (-1)-2 (-1)2

Maximum area = 6+4-2 = 8

Please follow the same steps for equation given in part (d)

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