Area of a cubic enclosed by 4 tangents

[flyingKangaroo]

Hey guys,

I have a cubic function f(x)=(x+1)(x+4)(x+6) and have 4 tangents drawn, 2 of which are parallel to each other, resulting in a parallelogram. I need to find the area enclosed by these tangents.

I was thinking about it and have concluded that the best method would be to find the points of intersection of the tangents with each other, then find the distance between the points., then use A=bh to find the area.

Is this the best way? Any other ideas? Any help would be great

Thanks,
flyingKangaroo

 

[Deleted member 4993]

Hey guys,

I have a cubic function f(x)=(x+1)(x+4)(x+6) and have 4 tangents drawn, 2 of which are parallel to each other, resulting in a parallelogram. I need to find the area enclosed by these tangents. What else do you know about the locations of these tangents?

 

[Pklarreich]

I was thinking about it and have concluded that the best method would be to find the points of intersection of the tangents with each other, then find the distance between the points. . I think you want the distance between the lines; that will be your 'height'.

 

[flyingKangaroo]

Okay, so the four tangents are:
y=10x+60
y=10x+9.191
y=-x=1
y=-x-10.481

I have the points that the four tangents intersect and the distance between these points. Since it is a parallelogram, the 'base' has a distance of 6.5 and a 'height' of 8.6. Using A=bh I get A=55.9. Is this correctly finding the area? Or is splitting the area into two triangles and using Heron's rule the correct way to go about this? If it is then I have no idea how to go about it...

Thanks for the help

 

[mmm4444bot]

Okay, so the four tangents are:
y=10x+60
y=10x+9.191
y=-x=1 It appears that this equation is supposed to be y = -x - 1; is that correct?
y=-x-10.481

[attachment=0:2ortiw5f]junk744.jpg[/attachment:2ortiw5f]

I'll call the height of the parallelogram the short thick line segment; then the long thick line segment is the base.

I did not confirm your measures for the height and base, but did you do the following?

The length of the base is found using the distance formula with the coordinates of the intersection points at each end.

The length of the height is found by first determining the equation of the line passing through the point at the right end of the base, perpendicular to the base. This line intersects the yellow tangent line, and those coordinates can be determined. Then the distance formula is used with the coordinates of the points at each end of the height.

A = h * b yields the correct area.

Cheers,

~ Mark

 

[flyingKangaroo]

flyingKangaroo wrote:Okay, so the four tangents are:
y=10x+60
y=10x+9.191
y=-x=1 It appears that this equation is supposed to be y = -x - 1; is that correct?
y=-x-10.481

Yes, sorry, the equation is y=-x-1

Thank you for the help. So, just checking, the area found using A=bh is the same as if I were to split the area into 2 triangles and use Heron's rule?

 

[mmm4444bot]

... So, just checking, the area found using A=bh is the same as if I were to split the area into 2 triangles and use Heron's rule?

Ah, yes. You will get the same area. (Excuse me if I missed that question earlier.)

Any geometric object that has area can be torn into as many pieces as you like without altering the aggregate area of the pieces (i.e, the whole). One difference is that having more than one piece may turn out to be more work.

Cheers,

~ Mark

BY THE WAY, I THINK HERON'S FORMULA ROCKS! I MIGHT VERY WELL USE IT EVEN IF IT IS MORE WORK SOMETIMES FOR NO OTHER REASON THAN THAT IT'S FUN.