Geometrical Applications of Calculus (2 Viewers)

[FDownes]

Oh man, another thread... Oh well. So, here's my problem;

Find the first and second derivatives of (5 - x)/(4x² + 1)²

Pretty simple, although I must be making a mistake in my working somewhere... Can anyone help?

 

[namburger]

First Derivative:
(12x^2 - 80x - 1) / (4x^2+1)^3
Second Derivative:
16 (12x^3 - 100x^2 - 3x + 5)/ (4x^2+1)^4

 

[FDownes]

Could you go through your working please? I need to figure out where I was making the mistake.

 

[namburger]

Just using the quotient rule in each one, if there's a mistake, it will most likely be due to the immense amount of algebra. Try simplifying as much as possible to make life easier
EDIT: did you get first derivative right or are you stuck on second?

 

[FDownes]

Stuck on the second.

EDIT: Nope, still can't get it right.

 

[namburger]

Stuck on the second.

y' = (12x^2 - 80x - 1) / (4x^2+1)^3
Let A = 4x^2+1

y'' = [(24x-80)A^3 - 3A^2.8x.(12x^2 - 80x - 1)]/A^6
=[(24x-80)A^3 - 24xA^2.(12x^2 - 80x - 1)]/A^6
=[8(3x-10)A^3 - 24xA^2(12x^2 - 80x - 1)]/A^6
=8[(3x-10)A - 3x(12x^2 - 80x - 1)]/A^4
=8[(12x^3-40x^2+3x-10) - 3(12x^2-80x-1)]/A^4
=8[-24x^3+200x^2+6x-10]/A^4
=-16 (12x^3 - 100x^2 - 3x + 5)/ (4x^2+1)^4

 

[FDownes]

Sorry, I should have checked earlier, but the textbook has a different answer to what you've got there. According to it, the first derivative is;

(20x² - 120x - 1)/(4x² + 1)⁴

And the second derivative is;

[-8(60x³ - 420x² - 9x + 15)]/(4x² + 1)⁵

 

[namburger]

Ok if you got the same answer as me for the first derivative, that is quite a coincidence if the answer is wrong. I'll redo it, and see if i get the same answer

y=(5 - x)/(4x^2 + 1)^2
let A = 4x^2 + 1

y'= [-A^2 - 2A.8x.(5-x)]/A^4
= [-A - 16x(5-x)]/A^3
= [- 4x^2- 1- 80x+ 16^2]/A^3
= (12x^2-80x-1)/(4x^2+1)^3

Therefore textbook is wrong

 

[FDownes]

It's weird, but yes, I did work out the same answer. Sorry about putting you through all this trouble.

 

[namburger]

lol NP
Just don't waste your time doing these questions with stupid values. Understand the process because you won't get these questions where algebra is a bitch and one mistake is fatal. The questions you get on diffrentiation should be usually simple, nothing to complex because they are looking if you are using the right process

 

[YannY]

" don't waste your time doing these questions with stupid values. Understand the process "

Shit, its the new age confucius.

Lols jk

 

[namburger]

new age confucius.

Lol, dam straight it is. =D

 

[tacogym27101990]

yeah it looks like the book is wrong

 

[FDownes]

Thanks guys, but it looks like I've run in to more troubles;

Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space.

So, I figure

= A = 1/2xy
= x² + y² = 8²
= y² = 64 - x²
= y = (64 - x²)^1/2
= A = 1/2x(64 - x²)^1/2
= A' = 1/2(64 - x²)^1/2 - x²/4(64 - x²)^-1/2

After that, my working falls apart. Can someone help me out?

EDIT: Man I'm bad at this subject....

Find any stationary points or inflexions on the curve y = x(x - 2)³

I just end up with stat points at x = 0 and 2, where they should be points of inflexion at x = 2 and 1, and a minimum at 1/2.

 

[Mark576]

y=x(x-2)³
y'=(x-2)³+3x(x-2)²=(x-2)²(4x-2)
y'=0 will give the stationary points:
For y'=0, x=2,1/2.
y''=2(x-2)(4x-2)+4(x-2)²=(x-2)(8x-4+4x-8)=(x-2)(12x-12)
From this we find x=1, is a point of inflection, and x=2 is a horizontal point of inflection [as it satisfies y'=0 and y''=0], which we can verify by observing the sign of y'' as it passes through x=2.
Testing x=1/2 for maximum or minimum:
y''=(1/2-2)(6-12) > 0 => x=1/2 is a minimum stationary point.

 

[FDownes]

Ugh... I'm such an idiot. Thanks.

Still stuck on the other questions though...

 

[namburger]

A' = 1/2(64 - x²)^1/2 - x²/4(64 - x²)<sup>-1/2

Ill assume the top is correct
let A' = 0
Multiply both sides by </sup>(64 - x²)<sup>1/2

(64-x^2)/2 - x^2/4 = 0
2(64-x^2)-x^2=0
128-3x^2=0
x=6.53(to 2.d.p.)
</sup>

 

[FDownes]

Hmm... The answer in the book is 16m², ergo my working for that question is probably wrong.

*sigh* This isn't good. I have a mathematics test on wednesday...

 

[namburger]

Hmm... The answer in the book is 16m², ergo my working for that question is probably wrong.

*sigh* This isn't good. I have a mathematics test on wednesday...

A = 1/2x(64 - x²)<sup>1/2

Sub in 6.53 and you get 15, which is quite close haha
</sup>

 

[FDownes]

Ugh... MORE help needed;

Find the greatest and least values of the function f(x) = 4x³ - 3x² - 18x in the domain -2 ≤ x ≤ 3.

Man I suck...