Post a New Question

MaTH!!!

posted by .

Let P(x,y) be a point on the curve y=16-x^2 and let A be the point directly below P on the x axis. If O is the origin, find the area and the perimeter of triangle OAP as functions of x.

I got Area=x(8-1/2x^2). Is that right?

And perimeter= x+16-x^2+ (square root 256-30x^2)

Did I do this question right?

  • MaTH!!! -

    Your area is correct

    But for the perimeter, ....
    let the hypotenuse by y
    y^2 = x^2 + (16-x^2)^2
    = x^2 + 256 - 32x^2 + x^4
    y = √(256 - 31x^2 + x^4)

    perimeter = x + √(256-31x^2+x^4) + 16- x^2

    testing:
    let P be (3, 7)
    then AP = 7
    OA = 3 and OP = √(9+49) = √58
    Perimeter = 3 + 7 + √58 = 10+ √58

    according to my formula
    Perimeter = 3 + √(256 - 31(9) + 81) + 16 - 9
    = 10 + √58

    Well, how about that?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question