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Need help with Differentiation!

OP

cool, thanks :smile:

well, i've just done it, and so am sure that the function is given by:

c = (t²/200)(200-t²)

dc/dt = (t/100)(200-t²) + (-2t)(t²/200)
dc/dt = 2t - (t³/100) - (t³/100)
dc/dt = 2t - (t³/50)

a) t=8, dc/dt = 16-10.24 = 5.76 g/L.s

b) t=10, dc/dt = 20 - 20 = 0

Reply 22

k@tie

10

mockel

well, i've just done it, and so am sure that the function is given by:

c = (t²/200)(200-t²)

dc/dt = (t/100)(200-t²) + (-2t)(t²/200)
dc/dt = 2t - (t³/100) - (t³/100)
dc/dt = 2t - (t³/50)

a) t=8, dc/dt = 16-10.24 = 5.76 g/L.s

b) t=10, dc/dt = 20 - 20 = 0

hence why mine wasn't working coz I was doing it the other way *screams*
I think this beats your best post btw....hands down.

Reply 23

m:)ckel

14

Katie Heskins

I think this beats your best post btw....hands down.

no way! did you even see my best post!?

Reply 24

k@tie

10

mockel

no way! did you even see my best post!?

I lied no....I just saw u mention it!! what was ur best post?!

Reply 25

m:)ckel

14

Katie Heskins

I lied no....I just saw u mention it!! what was ur best post?!

the one in my sig

:tsr:

Reply 26

k@tie

10

mockel

the one in my sig

:tsr:

ur right that looks really pretty.
ru doing maths next year?

Reply 27

m:)ckel

14

Katie Heskins

ur right that looks really pretty.
ru doing maths next year?

yeah, hopefully
i'm guessing you are too (or maybe maths w/ economics?)

Reply 28

TheWolf

15

For the curve with equation y = 2x/1+x^2

(a) prove that dy/dx = 2(1-x^2)/(1+x^2)^2

if anyone could be bothered to do this, itd be cool :smile:

thanks in advance.

Reply 29

m:)ckel

14

TheWolf

For the curve with equation y = 2x/1+x^2

(a) prove that dy/dx = 2(1-x^2)/(1+x^2)^2

if anyone could be bothered to do this, itd be cool :smile:

thanks in advance.

use the quotient rule:

dy/dx = [(2)(1+x^2) - (2x)(2x)] / (1+x^2)^2

Expand and simplify:

dy/dx = (2 - 2x^2) / (1+x^2)^2
dy/dx = 2(1-x^2) / (1+x^2)^2

Reply 30

TheWolf

15

thanks! last one:

The curve with equation y=e^x(px^2 + qx + r) is such that the tangents at x=1 and x=3 are parallel to the x-axis. The point (0,9) is on the curve. Find the vaules of p,q,r.

Reply 31

m:)ckel

14

TheWolf

thanks! last one:

The curve with equation y=e^x(px^2 + qx + r) is such that the tangents at x=1 and x=3 are parallel to the x-axis. The point (0,9) is on the curve. Find the vaules of p,q,r.

after differnetiating, you should get:

dy/dx = e^x (px^2 + qx + 2xp + q + r)

if the lines x=1 and x=3 are tangents, parallel to the x-axis, their gradient must be 0, i.e dy/dx=0
Substituting this into the above equation, we get:

when x=1, 0 = e(3p+2q+r)
=> 3p+2q+r = 0 ------(1)

when x=3, 0 = e^3(15p+4q+r)
=> 15p+4q+r = 0 ------(2)

Now sub in (0,9) into the original equation:
9 = 1(0+0+r)
r = 9

Sub 'r' into the other equations giving

3p+2q = -9 -----(3)
(3)*2 => 6p+4q = -18 ------(4)

15p+4q = -9 -------(5)

(5) - (4) => 9p = 9
p = 1

sub 'p' into (3) => 2q = -9-3
q = -6

Need help with Differentiation!

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