MedVision ad

Integration (1 Viewer)

Joined
Jul 21, 2013
Messages
68
Gender
Undisclosed
HSC
2014
Hi,

Find y as a function of x, if y' = 1/4x and y= 1 and y = 1 when x = e2. What is the x intercept of this curve?

If someone could explain this to me, it would be great. Thank you.
 

AnimeX

Member
Joined
Aug 11, 2012
Messages
588
Gender
Male
HSC
N/A
Hi,

Find y as a function of x, if y' = 1/4x and y= 1 and y = 1 when x = e2. What is the x intercept of this curve?

If someone could explain this to me, it would be great. Thank you.
is that a typo?
 

SharkeyBoy

Member
Joined
Nov 15, 2012
Messages
180
Gender
Male
HSC
2013
y' = 1/4 x
y = 1/8 x^2 + c
1 = 1/8 e^4 + c
c = 1 - 1/8 e^4
therefore:
y = 1/8 x^2 - 1/8 e^4
to find x int, sub y =0
therefore x = plus/minus e^2
 

D94

New Member
Joined
Oct 5, 2011
Messages
4,423
Gender
Male
HSC
N/A
"y' = 1/4 x
y = 1/8 x^2 + c"

Thats differentiation, if u differentiate y' (dy/dx) with respect to x, then u get y'' (d^2 y/dx^2), not y
It's because SharkeyBoy interpreted the integrand as (1/4)x, whereas you interpreted it as 1/(4x). You are both correct, given your assumptions.

Also, this was the first of two identical threads created.
 

HeroicPandas

Heroic!
Joined
Mar 8, 2012
Messages
1,547
Gender
Male
HSC
2013
It's because SharkeyBoy interpreted the integrand as (1/4)x, whereas you interpreted it as 1/(4x). You are both correct, given your assumptions.

Also, this was the first of two identical threads created.
oh right lol (and it he didnt differentiate aswell lol...)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top