Another math topic!

2009-02-11 10:19:20

#1

SteikeTa: Member; +153|7031|Norway/Norwegen/ Norge/Noruega

The function f is given by f(x) = e^2x-1

- Find the equation for the tangent (don't know if that is the correct word in English) in the spot (1, e)

The answer is: y= 2ex-e

Could anyone show me how to do this? My guess is to first ... ehm, derivate (??) it? You know:

f(x) = e^2x-1

f ' (x)= 2x * e^2x-1 <---- if that is correct, I don't know how to proceed further.

2009-02-11 10:50:11

#2

presidentsheep: Back to the Fuhrer; +208|6245|Places 'n such

f(x) = e^2x-1
differentiate to find the gradient
f'(x) = 2x-1(e¹)
multiply out the bracket bit to get f(x) = 2ex-e
then to get the gradient you plug in the x value (1)

gives y = 2ex-e

I think.

Last edited by presidentsheep (2009-02-11 10:50:37)

I'd type my pc specs out all fancy again but teh mods would remove it. Again.

2009-02-11 11:32:15

#3

SteikeTa: Member; +153|7031|Norway/Norwegen/ Norge/Noruega

Hmmm, not really sure about that.

2009-02-11 11:46:23

#4

a_member: Member; +13|6817|London, UK

First you differentiate to find the gradient:
f'(x)=2*e^(2x-1) (not 2x*e^(2x-1))
At x=1 the gradient is 2*e^(2-1)=2e

The tangent at x=1 is therefore a straight line with gradient 2*e
So we have a straight line of the form y=2ex+c where c is a constant

Sub in x=1, y=e to get
e=2e+c
c=-e

So the tangent is y=2ex-e

2009-02-11 12:02:01

#5

Jenspm: penis; +1,716|7016|St. Andrews / Oslo

a_member wrote:
First you differentiate to find the gradient:
f'(x)=2*e^(2x-1) (not 2x*e^(2x-1))
At x=1 the gradient is 2*e^(2-1)=2e

The tangent at x=1 is therefore a straight line with gradient 2*e
So we have a straight line of the form y=2ex+c where c is a constant

Sub in x=1, y=e to get
e=2e+c
c=-e

So the tangent is y=2ex-e

this.

@SteikeTa: "Kjerneregel" for derivation.

Set u=2x-1

f'(x) = (e^u)' * (2x-1)'
f'(x) = e^u * 2 (as a_member said)

https://static.bf2s.com/files/user/26774/flickricon.png

2009-02-11 12:26:15

#6

SteikeTa: Member; +153|7031|Norway/Norwegen/ Norge/Noruega

a_member wrote:
First you differentiate to find the gradient:
f'(x)=2*e^(2x-1) (not 2x*e^(2x-1))
At x=1 the gradient is 2*e^(2-1)=2e

The tangent at x=1 is therefore a straight line with gradient 2*e
So we have a straight line of the form y=2ex+c where c is a constant

Sub in x=1, y=e to get
e=2e+c
c=-e

So the tangent is y=2ex-e

Thanks, I get it now.

2009-02-11 16:24:25

#7

Vub: The Power of Two; +188|6778|Sydney, Australia

Jenspm wrote:
a_member wrote:
First you differentiate to find the gradient:
f'(x)=2*e^(2x-1) (not 2x*e^(2x-1))
At x=1 the gradient is 2*e^(2-1)=2e

The tangent at x=1 is therefore a straight line with gradient 2*e
So we have a straight line of the form y=2ex+c where c is a constant

Sub in x=1, y=e to get
e=2e+c
c=-e

So the tangent is y=2ex-e
this.

@SteikeTa: "Kjerneregel" for derivation.

Set u=2x-1

f'(x) = (e^u)' * (2x-1)'
f'(x) = e^u * 2 (as a_member said)

Which_member?
Sorry, had to do it.
But yes, a_member is on the ball. (which_member damn it!)

Last edited by Vub (2009-02-11 16:24:54)

2009-02-13 14:00:37

#8

Sheen1101: Member; +12|6903

Does anybody know how to do this problem?

https://img401.imageshack.us/img401/7868/math5kk0.jpg

I know how to do part (b) which is:
s'(t) = 2t - 8
s'(4) = 2(4) - 8
s'(4) = 0

Though I do not quite know how to do part (a). I've tried plugging in the numbers in for 2 t's and solving for s(t) but did not work. So can someone help me? I need to finish this because this homework is due tonight.

2009-02-13 14:18:44

#9

Finray: Hup! Dos, Tres, Cuatro; +2,629|6072|Catherine Black

Maths, you amerifags.

2009-02-13 14:20:49

#10

Ultrafunkula: Hector: Ding, ding, ding, ding...; +1,975|6757|6 6 4 oh, I forget

Finray wrote:
Maths, you amerifags.

Hush. Your steeringwheel is on the wrong side.

2009-02-13 14:33:57

#11

Sheen1101: Member; +12|6903

Ultrafunkula wrote:
Finray wrote:
Maths, you amerifags.
Hush. Your steeringwheel is on the wrong side.

um......okay? That really didn't help me with my homework.......

2009-02-13 14:35:40

#12

Ultrafunkula: Hector: Ding, ding, ding, ding...; +1,975|6757|6 6 4 oh, I forget

Sheen1101 wrote:
Ultrafunkula wrote:
Finray wrote:
Maths, you amerifags.
Hush. Your steeringwheel is on the wrong side.
um......okay? That really didn't help me with my homework.......

Makes you think

2009-02-13 14:36:37

#13

DesertFox-: The very model of a modern major general; +796|6968|United States of America

Average Value Theorem, I would think

2009-02-14 02:23:00

#14

B-Scimitar: Defeating your warriors.; +116|6693|Espoo, Finland

Sheen1101 wrote:
Does anybody know how to do this problem?

http://img401.imageshack.us/img401/7868/math5kk0.jpg

I know how to do part (b) which is:
s'(t) = 2t - 8
s'(4) = 2(4) - 8
s'(4) = 0

Though I do not quite know how to do part (a). I've tried plugging in the numbers in for 2 t's and solving for s(t) but did not work. So can someone help me? I need to finish this because this homework is due tonight.

Try with ( s'(t) - s'(t₀) ) / ( t - t₀ ), where t₀ is the time at the beginning and t at the end of the interval.

2009-02-14 02:52:37

#15

nukchebi0: Пушкин, наше всё; +387|6608|New Haven, CT

All you do is find the area under the velocity graph on each of those intervals, and then divide it by the length of the interval. This is, as noted, the Mean Value Theorem.

If v(t) = velocity, then the answer would be done as V(larger number) - V(smaller number) / (larger number - smaller number), where V(t) is the integral of v(t). As s'(t) = v'(t), V(t) = s(t). Thusly, you don't even need to integrate. Just use s(t) in the manner described above.

2009-03-06 17:16:13

#16

TheunforgivenII: Member; +12|6566

I need help with this problem......

https://img21.imageshack.us/img21/5907/90582218.jpg

2009-03-06 18:16:19

#17

DrunkFace: Germans did 911; +427|6965|Disaster Free Zone

Ultrafunkula wrote:
Finray wrote:
Maths, you amerifags.
Hush. Your steeringwheel is on the wrong side.

On the right side faggot.

2009-03-06 21:17:04

#18

TheunforgivenII: Member; +12|6566

um..... yeah well I still need help with my Calculus problem

2009-03-06 21:54:51

#19

The A W S M F O X: I Won't Deny It; +172|5968|SQUID

2009-03-07 00:16:42

#20

mcminty: Moderating your content for the Australian Govt.; +879|7005|Sydney, Australia

TheunforgivenII wrote:
um..... yeah well I still need help with my Calculus problem

How did you get answer (e) correct? Like.. what equations and stuff have you come up with?

2009-03-07 01:39:20

#21

mcminty: Moderating your content for the Australian Govt.; +879|7005|Sydney, Australia

Theunforgiven, I haven't checked the answers this gives but..

1. In 20 minutes, where P₀ is the initial population of 55;

P(t) = P₀ x 2

2. In an hour, with 3 lots of 20 minutes..

P(t) = P₀ x 2 x 2 x 2 = P₀ x 2³ = 55(2)³

3. So, generally, for when you have t hours:

P(t) = 55(2)^3t

Now, it asks for "k" in part (a), so I'm guessing that relates to:

P(t) = P₀e^kt as the general formula for exponential growth.

Equate the two and you get:

e^kt = 2^3t

Reducing to k = 3ln2 (ln is the natural logarithm..)

Check this by putting it into P(t) = P₀e^{3ln2 x t}, and plugging in some values. I did it and it works out..

Now you can find the rates of change by determining dP/dt.

2009-03-07 09:51:31

#22

TheunforgivenII: Member; +12|6566

mcminty wrote:
Theunforgiven, I haven't checked the answers this gives but..

1. In 20 minutes, where P₀ is the initial population of 55;

P(t) = P₀ x 2

2. In an hour, with 3 lots of 20 minutes..

P(t) = P₀ x 2 x 2 x 2 = P₀ x 2³ = 55(2)³

3. So, generally, for when you have t hours:

P(t) = 55(2)^3t

Now, it asks for "k" in part (a), so I'm guessing that relates to:

P(t) = P₀e^kt as the general formula for exponential growth.

Equate the two and you get:

e^kt = 2^3t

Reducing to k = 3ln2 (ln is the natural logarithm..)

Check this by putting it into P(t) = P₀e^{3ln2 x t}, and plugging in some values. I did it and it works out..

Now you can find the rates of change by determining dP/dt.

thanks mcminty....I got letter (e) correct by doing 20(log(20000/55)) / (log(2)) = 170.127 min => 2.8345....hours. It was just luck that I got it right...I had no idea what I was doing at first.

Last edited by TheunforgivenII (2009-03-07 09:51:55)

2009-03-07 17:13:14

#23

mcminty: Moderating your content for the Australian Govt.; +879|7005|Sydney, Australia

No worries

2009-04-01 18:08:34

#24

Sheen1101: Member; +12|6903

I have two calculus problems that I can't seem to get right and I don't know what I'm doing wrong.....

1) Find the most general antiderivative of the function. Use C for any needed constant.
f(x) = 6x⁶ - 2x⁴ + 13x²

for this one I did 6(1/7)x⁷ + 2(1/5)x⁵ + 13(1/3)x³ and got .857x⁷ + .4x⁵ + 4.33x³ + C.....but this is wrong..

2) Find the most general antiderivative of the function. Use C for any needed constant.
f(x) = 5(√x) + 4cos(x)

on this one I started 5(1/(1+1/2))(x^1+(1/2)) + 4sinx + C
= 5(2/3)(x^3/2) + 4sinx + C
= 10/3x^3/2 + 4sinx + C which is also wrong....

So I need help guys....

Last edited by Sheen1101 (2009-04-01 18:08:55)

2009-04-01 19:51:40

#25

Sheen1101: Member; +12|6903

Sheen1101 wrote:
I have two calculus problems that I can't seem to get right and I don't know what I'm doing wrong.....

1) Find the most general antiderivative of the function. Use C for any needed constant.
f(x) = 6x⁶ - 2x⁴ + 13x²

for this one I did 6(1/7)x⁷ + 2(1/5)x⁵ + 13(1/3)x³ and got .857x⁷ + .4x⁵ + 4.33x³ + C.....but this is wrong..

2) Find the most general antiderivative of the function. Use C for any needed constant.
f(x) = 5(√x) + 4cos(x)

on this one I started 5(1/(1+1/2))(x^1+(1/2)) + 4sinx + C
= 5(2/3)(x^3/2) + 4sinx + C
= 10/3x^3/2 + 4sinx + C which is also wrong....

So I need help guys....

nobody is going to help me

BF2S Forums Another math topic!

a_member wrote:

a_member wrote:

Jenspm wrote:

a_member wrote:

Finray wrote:

Ultrafunkula wrote:

Finray wrote:

Sheen1101 wrote:

Ultrafunkula wrote:

Finray wrote:

Sheen1101 wrote:

Ultrafunkula wrote:

Finray wrote:

TheunforgivenII wrote:

mcminty wrote:

Sheen1101 wrote:

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