Help! (1 Viewer)

[cook E]

I need help on this question.

Find the equation of the tangent to the parabola y=2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola,the tangent line and the x-axis is revolved around the x-axis.

I found the Eqn of the tangent : y= 4x-2
and the Point of Intersection: (1/2,0)

But I can't seem to get the Volume correct.
The answer is 2pi/3 units cubed.

Can someone help me ?

 

[Bored Of Fail 2]

I need help on this question.

Find the equation of the tangent to the parabola y=2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola,the tangent line and the x-axis is revolved around the x-axis.

I found the Eqn of the tangent : y= 4x-2
and the Point of Intersection: (1/2,0)

But I can't seem to get the Volume correct.
The answer is 2pi/3 units cubed.

Can someone help me ?

where did you get this question from??, pretty sure its a bad question and has a mistake

 

[cook E]

where did you get this question from??, pretty sure its a bad question and has a mistake

3 Unit Mathematics Book 1 S.B Jones and K.E Couchman

 

[Bored Of Fail 2]

you did get the tangent line correct

but you did make a mistake when you found the point of intersection

solve y= 2x^2 and y= 4x-2 simutaneously

they both equal y, so set them equal to each other

2x^2 = 4x -2
2x^2 -4x +2 =0
2 ( x^2 -2x +1 ) =0
2 ( x-1)^2 =0

x=1 is the only intersection

there is no well defined region which they want use to rotate , the line doesnt intersect the parabola twice

 

[Bored Of Fail 2]

does it specify something like "being rotated about the POSITIVE" x axis " ??

 

[cook E]

you did get the tangent line correct

but you did make a mistake when you found the point of intersection

solve y= 2x^2 and y= 4x-2 simutaneously

they both equal y, so set them equal to each other

2x^2 = 4x -2
2x^2 -4x +2 =0
2 ( x^2 -2x +1 ) =0
2 ( x-1)^2 =0

x=1 is the only intersection

there is no well defined region which they want use to rotate , the line doesnt intersect the parabola twice

But it says find the point of intersection with the x-axis

so wouldn't it be

4x-2 = 0
x= 2/4
x= 1/2

And the answers also say the same.

does it specify something like "being rotated about the POSITIVE" x axis " ??

Nope, it just says whatever is in the question.

 

[Bored Of Fail 2]

But it says find the point of intersection with the x-axis

so wouldn't it be

4x-2 = 0
x= 2/4
x= 1/2

And the answers also say the same.





Nope, it just says whatever is in the question.

ahh yeh I just read the question again and figured that lol

 

[cook E]

lol, yeh this question is driving me insane

 

[Bored Of Fail 2]

ahh i got the picture now

so y= 2x^2 is above the line

so the formula is pi integral (y2)^2 - (y1)^2 dx

and the limits of the integral are 1 to 1/2 ( the 1 comes from solving the parabola and line simultaneously )
so answer= pi intergal [ 1 .. 1/2 ] ( 2x^2 )^2 - ( 4x -2)^2 dx

 

[Bored Of Fail 2]

ahh not sure thats it either :|

 

[cook E]

ahh not sure thats it either :|

yeah thats not it =[

 

[Bored Of Fail 2]

The region which im pretty sure they are interested in is the area 0<=x<=1

there region from 0<=x<=1/2 is under the parabola y=2x^2 , and the area between 1/2<=x<=1 consist of a region between the parabola and the line ( with the parabola above the line )

so answer should equal pi integral [ 0..1/2] ( 2x^2 ) ^2 dx + pi integral [ 1/2 .. 1] ( 2x^2 )^2 - ( 4x-2)^2 dx

that should give the answer, but when I do that I come up with 2pi/15 ( which could quite possibly be the answer, its not a extremely weird fraction) I think the answer given is wrong

 

[cutemouse]

I get 2pi/15 too