Question
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?
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?
Answers
Reiny
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?
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?