Finding area by integration problem

[lionely]

Homework Statement

The parabola y = 6x - x^2 meets the x-axis at O and A. The tangent at O and A meet at T. Show that the curve divides the area of the triangle OAT into two parts in the ratio 2:1.

Homework Equations

The Attempt at a Solution

Here is my working with my sketch. So I thought I should find the area bounded the Tangent at either A or O and the x-axis to find the area of the triangle . So I integrated from x=6 to x=0 and got 108m^2.

I also wanted to check it another way so I found the coordinates of T(3,18) and used that to find the area of an isosceles triangle the normal way, but I got 54m^2. So I am a bit lost now.

And just to check something I found the area bound by the curve and the x-axis from x = 6 to x =0 and got 36m^2.
I don't know what to do with that 36m^2 yet.

I know 108:54
is 2:1... but I don't think that's the correct method.

 

[Nathanael]

So you have two areas, you know that one of them is twice the other, and you also know what their sum is.

Try to write this information with equations.

 

[lionely]

Hold on I actually have three areas there. Should I ignore the 54m^2?

 

[Nathanael]

The 54m^2 is a mistake, somehow you only accounted for half of the triangle
(I don't know where you went wrong because you didn't show how you got it).

 

[lionely]

Oh okay sorry. Area bounded the curve I'll call it C and area of Triangle AOT. C+AOT = 36 + 108 = 144

and AOT = 3C
108 = 3(36).. Hmmm so what should I do now?

 

[Mark44]

So you have two areas, you know that one of them is twice the other, and you also know what their sum is.

Try to write this information with equations.

The OP doesn't "know" this about the two areas; he has to show this. Here's from post #1.

Show that the curve divides the area of the triangle OAT into two parts in the ratio 2:1.

 

[SteamKing]

The Attempt at a Solution

Here is my working with my sketch. So I thought I should find the area bounded the Tangent at either A or O and the x-axis to find the area of the triangle . So I integrated from x=6 to x=0 and got 108m^2.

Exactly what did you integrate here? Please show your work.

I also wanted to check it another way so I found the coordinates of T(3,18) and used that to find the area of an isosceles triangle the normal way, but I got 54m^2. So I am a bit lost now.

54 is the correct area for a triangle with base = 6 and altitude of 18.

And just to check something I found the area bound by the curve and the x-axis from x = 6 to x =0 and got 36m^2.

That's the correct area under the parabola between x = 0 and x = 6.

I don't know what to do with that 36m^2 yet.

Hold on to it and don't lose it.

I know 108:54
is 2:1... but I don't think that's the correct method.

If the curve divides the area of triangle OAT into two parts with ratio 2:1, then this means that 2:1 represents the ratio of the area of the triangle which is not under the parabola to the area which is under the parabola.

You've found the area under the parabola. What is the area under the triangle OAT but also not under the parabola? What is the ratio of these two areas?

 

[SteamKing]

Oh okay sorry. Area bounded the curve I'll call it C and area of Triangle AOT. C+AOT = 36 + 108 = 144

and AOT = 3C
108 = 3(36).. Hmmm so what should I do now?

I don't know what you are doing here, but you are getting colder, very much colder.

 

[SteamKing]

The 54m^2 is a mistake, somehow you only accounted for half of the triangle
(I don't know where you went wrong because you didn't show how you got it).

You should show your work.

 

[lionely]

Exactly what did you integrate here? Please show your work.

I integrated the equation of the tangent at A ; y = 36 -6x from x = 6 to x = 0. Also I integrated the equation of the tangent at O ; y = 6x for the same interval. Got 108m^2 in both cases. I was doing this because I thought the areas under either one of these curves would be the area of the triangle.

Okay, so the curve split the area of the triangle into an Area above the curve and one below it.

area of the triangle under the curve would be 36m^2.
Area above the curve would be = Area of triangle - Area below the curve = 54- 36 = 18m^2

Area of Triangle below the curve:Area of Triangle Above the Curve
36:18
2:1.

Thank you.
Hold on I will take a picture of my other work to show you what I did.
But why is the integral that gave me 108m^2 wrong?

 

[Nathanael]

But why is the integral that gave me 108m^2 wrong?

Because you're integrating under a single tangent over the whole interval. (You're taking the area of a large right triangle.) You would want to integrate under one tangent for part of the interval and the other tangent for the other part.

The OP doesn't "know" this about the two areas; he has to show this. Here's from post #1.

The OP knows what they are trying to show, my point was just that they should say it with equations...

You should show your work.

Oops.

 

[SteamKing]

I integrated the equation of the tangent at A ; y = 36 -6x from x = 6 to x = 0. Also I integrated the equation of the tangent at O ; y = 6x for the same interval. Got 108m^2 in both cases. I was doing this because I thought the areas under either one of these curves would be the area of the triangle.

But why is the integral that gave me 108m^2 wrong?

That's the tricky thing about this triangle. The two sides of the triangle are expressed by different equations, so you can't integrate the equation for one tangent over the entire interval and expect to get the correct area under the triangle.

 

[lionely]

Because you're integrating under a single tangent over the whole interval. (You're taking the area of a large right triangle.) You would want to integrate under one tangent for part of the interval and the other tangent for the other part.

I just tried it over when I integrate the tangent at A from x=6 to x =3 I get 27 and when I integrate the tangent at O from x=3 to x= 0 I also get 27. So I see the 54m^2.
But I still don't fully understand why if I use the tangent at A and use the whole interval, I get 108m^2. Do you have another example of this?

 

[lionely]

That's the tricky thing about this triangle. The two sides of the triangle are expressed by different equations, so you can't integrate the equation for one tangent over the entire interval and expect to get the correct area under the triangle.

Ohhhhhhhh. Oh okay

 

[Mark44]

The OP doesn't "know" this about the two areas; he has to show this. Here's from post #1.

The OP knows what they are trying to show, my point was just that they should say it with equations...

But if you write what you're supposed to prove as an equation, you're tacitly assuming that the relationship holds, not showing it.