Can someone check and/or correct this Pre-Cal functions problem?

Question

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.

Answer 1

You have the right answer. But this problem is very easy to do if you realize what things look like. Draw it out.

The distance between the point on the curve (x,y) to the y axis (0, y) is x (thats one base)

The distance between the origin and (0, y) is x³
This is because (x, x³) has to be at the same height as (0, y) for it to be a right triangle.

Thus x(x³) /2 = x^4 / 2


I just realized it might also be possible to form a right triangle if the (0, y) point slides upward enough to make the right angle the one off the y = x³ function. If this is the case then you'd need to know the value of x or y to solve...

Answer 2

You can see that all your effort to find the length of the hypotenuse, and then use that to find the length of one of the sides, was the long way 'round. As others have said, it is much easier to find the "base" and "height" of the triangle, and use

A = 1/2 * b * h

Scott is wrong to think that you can make the y value so large that the right angle will be on the function line. The right angle is always on an axis. The other two points of the triangle are at the origin and on the function line.

Answer 3

I think this problem is a little simpler...you need to realize that one of the vertex of the triangle coincides with the y = x^3 equation. Using this fact, you know now the points for the right triangle are as follows:

(0,0) origin
(0,y)
and
(x,y)

BUT because this is a right triangle, the two y values of the triangle MUST be the same.

so y = x^3

Points again are:

(0,0)
(0,x^3)
(x,x^3)
Area of triangle is 1/2 (base)*(height)

Area = 1/2 x(y)
= 1/2 (x)(x^3) or 1/2 x^4

Good luck!

Answer 4

You are correct, but you made the problem too hard. The triangle is a right triangle with a height of x^3 and a base of x. This makes the area 1/2 base x height or 1/2 x^4. If you graph it out, you will see what I mean. Still, you got the correct answer, so congrats.

Answer 5

Something doesn't add up.

y = x^3 is not linear because its derivative is not constant.

I think you don't have a right triangle, but some curved polygon which requires an integral to calculate.

Answer 6

got see some of the answers these people will give you...