A few questions i cant get (1 Viewer)

[Mr Chi]

1) find the equation of the tangent to the curve xsquared = -2y at the point (4,-8) . Tangent meets directrix at M. Find M's cordinates.

2) ( 3 4x + 1/ 2x^2 +x+1 (evaluate to 1 decimal place)
)0

3) find the volume of the solid formed when the curve y= e^-x +1 is rotated about the x axis from x=1 to x=2, correct to 1 dp.

working out would help me alot thanks.

 

[m0ofin]

Oh I'm not sure if I've done it right, I'm a bit rusty (well, okay, maybe heaps rusty) at it but I hope it helps you Mr Chi, provided that it's right though.


EDIT// Oops, I just realised my answer was -ve

 

[Shady01]

1) differntiate x^2=-2y
y'= -x or osmthin like that (doin this in head btw)
then fer x sub in 4
therefore m= -4
y-y1=m(x-x1)
= y+8=-4(x-4)
and there u go also i cant read question 2 and hav no idea bout 3

 

[abcd9146]

question 2
i got 25.3 for the for answer (but it might be wrong, if im right ill post what i did)

question 3
i got 56.4u³, it thst right?

 

[o0_jolie_0o]

V = π∫y^2
= π∫(e^-x + 1)^2
= π∫(e^-2x + 2e^-x + 1)
= π[-1/2e^-2x - 2e^-x + x] (sub in x=2 and x = 1)
= π{ [ -1/2e^-4 - 2e^-2 + 2] - [-1/2e^-2 - 2e^-1 + 1]}
= 1.720... - 0.19657...
= π[1.52343....]
= 4.8 u^3 (1 d.p.)

 

[SoulSearcher]

1) differntiate x^2=-2y
y'= -x or osmthin like that (doin this in head btw)
then fer x sub in 4
therefore m= -4
y-y1=m(x-x1)
= y+8=-4(x-4)
and there u go also i cant read question 2 and hav no idea bout 3

To finish this part off, you get the equation 4x + y - 8 = 0
since the equation of the directrix is y = 1/2, substitute that value of y into the equation of the tangent to find out the x-coordinate.
so
4x + 1/2 - 8 = 0
8x + 1 - 16 = 0
8x = 15
x = 15/8 therefore coordinates of M is (15/8, 1/2)

I dont really get your second question.

 

[SoulSearcher]

3) find the volume of the solid formed when the curve y= e^-x +1 is rotated about the x axis from x=1 to x=2, correct to 1 dp.

y = e^-x + 1
y² = (e^-x + 1)²

therefore the integral is

π ∫ y^2 dx (2 --> 1)
= π ∫ (e^-x + 1)² dx (2 --> 1)
= π [ { -(e^-x +1)³/3 } + x ] (2 --> 1)
= -π [ { (e^-x +1)³/3 } - x ] (2 --> 1)
= -π [ { [ (e^-2 +1)³/3 ] - 2 } - { [ e^-1 +1)³/3 ] - 1 }
= -π [ -1.512 + 0.147 ]
= -π * -1.365
= 4.3 u³ to 1 dec. pl.

 

[Mountain.Dew]

y = e^-x + 1
y² = (e^-x + 1)²

therefore the integral is

π ∫ y^2 dx (2 --> 1)
= π ∫ (e^-x + 1)² dx (2 --> 1)
= π [ { -(e^-x +1)³/3 } + x ] (2 --> 1)
.....
= 4.3 u³ to 1 dec. pl.

how did u get that step in?

i dont think that u're allowed to integrate like that ==> the squared function must be linear to be able to integrate that way.

 

[Mountain.Dew]

Oh I'm not sure if I've done it right, I'm a bit rusty (well, okay, maybe heaps rusty) at it but I hope it helps you Mr Chi, provided that it's right though.


EDIT// Oops, I just realised my answer was -ve

ur answer isnt negative....i think thats the right answer by the looks of it.

 

[pLuvia]

how did u get that step in?

i dont think that u're allowed to integrate like that ==> the squared function must be linear to be able to integrate that way.

Just expand it out more correct