help on calculus

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jones001

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Nov 12, 2006
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1) A curve has equation y = (x^2) / (2x + 1)

. . .i) Find dy/dx.
. . . ..Hence, find the coordinates of the stationary
. . . ..points on the curve.

. . .ii) You are given that (d^2y)/(dx^2) = 2 / (2x + 1)^3
. . . . .Use this information to determine the nature
. . . . .of the stationary points in part (i).

. . .iii) Show that the substitution t = 2x + 1 converts
. . . .. .the integral int [ (x^2) / (2x + 1) ] dx to the
. . . .. .integral (1/8) int [ t + (1/t) - 2] dt

. . . .. .Hence, find the definite integral
. . . .. .int[x=0,1] [ (x^2) / (2x + 1) ] dx

Some help would be appreciated. Thank you.
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Edited by stapel -- Reason for edit: reformatting for size
 
part (i) ... pretty basic, just use the quotient rule to find dy/dx. set dy/dx = 0 to find the x-values of the "stationary" points, then substitute those x-values back into the original function to find the y-values.

part (ii) ... substitute the x-values of the "stationary" points into the second derivative ... the sign of the second derivative value tells you whether y has a max, min (or neither) at each "stationary point". Your text should say something regarding this procedure ... it's called the second derivative test.

part (iii) ... if t = 2x+1, then ...
1. dt = 2 dx or dt/2 = dx
2. x = (t-1)/2, so x<sup>2</sup> = (t-1)<sup>2</sup>/4

doing the substitution,
x<sup>2</sup>/(2x+1) dx = (t-1)<sup>2</sup>/(4t) * (dt/2)
a little algebra will get it looking like what you need.

finally, evaluate the definite integral using the fundamental theorem of calculus (FTC) ... reset your limits of integration to t-values from x-values using t = 2x+1.
 
Why did you square both sides of the equation x = (t - 1)/2 ? Or are you referring that to the Quotient Rule :?
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Edited by stapel -- Reason for edit: spelling, punctuation, etc
 
jones001 said:
Why did you square both sides of the equation x = (t - 1)/2 ?
Click to expand...
Probably in order to substitute, as he mentions in the next line.

jones001 said:
Or are you referring that to the Quotient Rule
Click to expand...
I'm not sure what is meant by the above, but since the Quotient Rule is related to differentiation and you're supposed to be integrating, the tutor probably isn't using the Quotient Rule. He's probably suggesting that you follow the step-by-step instructions given in the exercise, starting by by squaring and substituting. Now you just need to rearrange what he's given you, to get the integral to match what they're looking for.

If you get stuck in this process, please reply showing all of your work and reasoning. Thank you.

Eliz.
 
done part i and ii but iii? wat i no is that dx=dt/2 but y cant u just have x=0.5(t-1) rather than squaring
 
no dont worry i seen wat hes done. hes squared it so that i have the x substitution and the x^2 substitution
 
t = 2x+1 ... x = (t-1)/2 ... x<sup>2</sup> = (t-1)<sup>2</sup>/4
dt = 2dx ... dx = dt/2

substitute for x<sup>2</sup>/(2x + 1) dx ...

(t - 1)<sup>2</sup>/(4t) * dt/2 =

(1/8)(t<sup>2</sup> - 2t + 1)/t dt =

(1/8)(t - 2 + 1/t) dt

reset limits of integration ...
x = 0, t = 1
x = 1, t = 3

antiderivative is (1/8)[t<sup>2</sup>/2 - 2t + ln(t)]

use the FTC to evaluate ...

(1/8)[(9/2 - 6 + ln(3)) - (1/2 - 2 + 0)] = ln(3)/8
 
x<sup>2</sup> = (t - 1)<sup>2</sup>/4

2x+1 = t

... divide (t - 1)<sup>2</sup>/4 by t
 
y do u need to take the ln when u can just sub in the t values straight away...u get 1/6
 
i dont think u r meant to integrate the equation just found because the integral as limits, therefore all i need to do is sub the t values straight in, right?
 
y do u need to take the ln when u can just sub in the t values straight away...u get 1/6
Click to expand...

??? where in the world do you get 1/6 ???

you end up with an integrand of (1/8)(t^2 - 2t + 1)/t = (1/8)(t - 2 + 1/t)

You have to use the natural log because the antiderivative of 1/t is ln(t).
 
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