SunDun111
Badges: 2
Rep:
?
#1
Report Thread starter 5 years ago
#1
I don't understand the question where you have a equation and your expected to find the coordinates or the stationary point out of nothing? Like for example 3 root x - 5x +1, I can differentiate this but I csnt find the coordinates how do I do it? I can't solve it
0
reply
samb1234
Badges: 13
Rep:
?
#2
Report 5 years ago
#2
(Original post by SunDun111)
I don't understand the question where you have a equation and your expected to find the coordinates or the stationary point out of nothing? Like for example 3 root x - 5x +1, I can differentiate this but I csnt find the coordinates how do I do it? I can't solve it
Stationary points are where dy/DX is 0

Posted from TSR Mobile
0
reply
aoxa
Badges: 11
Rep:
?
#3
Report 5 years ago
#3
At stationary points, dy/dx = 0. From there, you rearrange the differential to get the x-coordinate of the stationary point. You sub this value back into the original equation to find the y coordinate of the stationary point.
0
reply
SunDun111
Badges: 2
Rep:
?
#4
Report Thread starter 5 years ago
#4
(Original post by aoxa)
At stationary points, dy/dx = 0. From there, you rearrange the differential to get the x-coordinate of the stationary point. You sub this value back into the original equation to find the y coordinate of the stationary point.
That's just it I can do dy/dx but I can't solve it when it's equal to 0, I don't understand how..
0
reply
aoxa
Badges: 11
Rep:
?
#5
Report 5 years ago
#5
(Original post by SunDun111)
That's just it I can do dy/dx but I can't solve it when it's equal to 0, I don't understand how..
What was the original equation of the curve, and what did you get when you differentiated it?
0
reply
SunDun111
Badges: 2
Rep:
?
#6
Report Thread starter 5 years ago
#6
(Original post by aoxa)
What was the original equation of the curve, and what did you get when you differentiated it?
When I dy\dx it is 3/2x minus to power of 1/2 - 5 = 0 I just can't work out coordinate
0
reply
SherlockHolmes
Badges: 13
Rep:
?
#7
Report 5 years ago
#7
(Original post by SunDun111)
When I dy\dx it is 3/2x minus to power of 1/2 - 5 = 0 I just can't work out coordinate
Rearrange for x and then substitute your x value into the original equation.
0
reply
aoxa
Badges: 11
Rep:
?
#8
Report 5 years ago
#8
(Original post by SunDun111)
When I dy\dx it is 3/2x minus to power of 1/2 - 5 = 0 I just can't work out coordinate
I presume you can rearrange it to x^(-1/2) = 10/3 - this is just adding the 5 across, multiplying by 2, then diving by 3 so you get the x term on it's own.

As you have the x term to the power of (negative) a half, this is the square root. To get rid of a square root, you square both sides to get x^-1 = 100/9

Can you rearrange that to get what x equals?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (49)
16.33%
I'm not sure (8)
2.67%
No, I'm going to stick it out for now (103)
34.33%
I have already dropped out (4)
1.33%
I'm not a current university student (136)
45.33%

Watched Threads

View All