# Differentiation - Could some one check my work

• Mar 12th 2011, 07:34 AM
djstar
Differentiation - Could some one check my work
hi, im quite new to calculus and find it taxing but enjoyable. ive been given 4 questions to answer and have had a good go at them. how ever other class mates have got completely different answers to me , could some one pleae just check over them to make sure im going in the correct direct. many thanks
Attachment 21125
• Mar 12th 2011, 08:12 AM
running-gag
Hi

If find the same answers but you haven't finished the last exercice
• Mar 12th 2011, 08:51 AM
djstar
Hi, thanks for the reply, as for the last question. What else do I have to do to it? Do I have to differentiate the previous differentiated formula and put my x values into that formula and use that answer as the x vale for the original formula? Many thanks.
• Mar 12th 2011, 10:07 AM
Ackbeet
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• Mar 12th 2011, 10:16 AM
running-gag
Quote:

Originally Posted by djstar
Hi, thanks for the reply, as for the last question. What else do I have to do to it? Do I have to differentiate the previous differentiated formula and put my x values into that formula and use that answer as the x vale for the original formula? Many thanks.

You have found the volume V(x) = (2-2x).(2-2x).(x)
You must determine the value of the piece to be cut out to give the maximum volume
This can be done by diffrentiating V(x)
• Mar 13th 2011, 03:33 AM
djstar
Thanks Ackbeet for the advice, ill remember that one.

ok so ive worked out the volume for the trough is

Volume = 4x3 – 8x2 +4x

[IMG]file:///C:/DOCUME%7E1/LIAMST%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]now i have found dy/dx = 12x2 – 8x +4

This has left me with a quadratic so i have put this into the quadratic equation

and have came up with two x answer's these are x=1 and x =0.33

do i put the 0.33 into the orignal equation 4(0.333 )– 8(0.332 )+4(0.33) to find the volume which i worked out as 0.592m2

Or do i have to differentiate the equation again

d2y/dx2 = 24x - 8

and substitue 0.33 into the euqation to give the answer -0.08

and then use this in the original formula 4(-0.083 )– 8(-0.082 )+4(-0.08) = -1.28 m2
• Mar 13th 2011, 03:44 AM
running-gag
V(x) = (2-2x).(2-2x).(x) = 4x(1-x)²
When differentiating V(x), you should not expand it
V'(x) = 4(1-x)²-8x(1-x) = 4(1-x)(1-3x)
It is easier to find the roots
V is defined for x between 0 and 1 therefore 1-x > 0
The sign of V'(x) is the same as 1-3x
• Mar 13th 2011, 05:28 AM
djstar
thanks for the reply running-gag , im sorry if im a pain for all the replies, but im really new to calculus and im still at secondary schoollearning it so all of these notations and formulas i can find confusing.

could i just check with you that the correct answer for the volume of the trough is 0.592 metres cubed?
• Mar 13th 2011, 06:42 AM
running-gag
You are right
The maximum volume is obtained when x=1/3 and V(1/3) = 16/27 = 0.593 m^3