Q: A right triangle has one vertex on the graph of y = x^3; x > 0, at (x , y), another at the origin, and the third on the positive y-axis at (0 , y). Express the area A of hte triangle as a function of x.
Here's my work:
Distance from (0,0) to (x,y) on y = x^3 equals d.
d = √[(x^2 + (x^3)^2]
d = √(x^2 + x^6)
Distance from (0,y) to (0,0) equals s.
s = √[(x^3)^2]
s = √(x^6)
s = x^3
Since d is the hypotenuse and s is one leg, to find the other leg (b for base), use:
(x^3)2 + b^2 = (√(x^2 + x^6))^2
x^6 + d^2 = x^2 + x^6
d^2 = x^2
d = x
To find the area, A, of the triangle, you perform
A = 1/2(bh)
A = 1/2(x)(x^3)
A = 1/2(x^4)
Right? There is a picture, with a graph. The right angle is formed at (0,y) and the hypotenuse is from (x,y) to (0,0). If you see anything wrong with my arithmetic, please point it out to me. Thank you, ahead of time.
2006-09-06
14:40:15
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6 answers
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asked by
Anonymous