I'm trying to understand calculus, I'm doing a basic but fast course and I haven't really covered much algebra and trigonometry so I'm struggling a bit. To check my answers and try to understand what is going on I decided to plot a graph.
I wasn't sure whether the graph was the y= or the dy/dx = so plotted two lines on the graph which look very different, the y= one is a pointy n shape and the dy/dx one is sort of u shape.
My assignment question is to find the coordinates of the stationary points and determine their nature. I did this and have the same answer as my friends.
The stationary points are on the dy/dx line (where the line crosses the x axis, is that right?). The coordinates I got are on the y= line.
I know I'm probably really dumb but can someone explain to me what it is all about please.
Please help explain calculus to me
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 16032012 21:39

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 16032012 21:44
(Original post by jami74)
I'm trying to understand calculus, I'm doing a basic but fast course and I haven't really covered much algebra and trigonometry so I'm struggling a bit. To check my answers and try to understand what is going on I decided to plot a graph.
I wasn't sure whether the graph was the y= or the dy/dx = so plotted two lines on the graph which look very different, the y= one is a pointy n shape and the dy/dx one is sort of u shape.
My assignment question is to find the coordinates of the stationary points and determine their nature. I did this and have the same answer as my friends.
The stationary points are on the dy/dx line (where the line crosses the x axis, is that right?). The coordinates I got are on the y= line.
I know I'm probably really dumb but can someone explain to me what it is all about please.
That'll give you your coordinates
EDIT:: Also if there is more than one solution when you solve for x, then you will have to do this for each solution of x to find the corresponding coordinate of y.Last edited by Lovin; 16032012 at 21:45.Post rating:1 
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 16032012 21:54
Imagine an x^2 graph
When the graph is going up it has a positive gradient (like going up a hill) similarly going down the hill has a negative gradient
At the top or bottom of the curve the gradient changes from +ve to ve (or ve to +ve depending on the graph) so that right at the turning point you will have a gradient of 0
dy/dx gives the gradient at any point of the curve so when the gradient = dy/dx is 0 you have a
turning point
stationary point
maximum
miniumPost rating:1 
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 16032012 23:49
(Original post by Lovin)
Stationary points occur when dy/dx=0 So differentiate to obtain dy/dx, put it as an equation which equals 0. Solve for x then put it back into the original equation to solve for y.
That'll give you your coordinates
EDIT:: Also if there is more than one solution when you solve for x, then you will have to do this for each solution of x to find the corresponding coordinate of y.
I then plugged my numbers into a formula b +or square root of b^2  4ac /2a and got two x = numbers.
Then I plugged my x = numbers into my original y= equation and got two sets of coordinates. So far this makes sense to me.
(Original post by TenOfThem)
Imagine an x^2 graph
When the graph is going up it has a positive gradient (like going up a hill) similarly going down the hill has a negative gradient
At the top or bottom of the curve the gradient changes from +ve to ve (or ve to +ve depending on the graph) so that right at the turning point you will have a gradient of 0
dy/dx gives the gradient at any point of the curve so when the gradient = dy/dx is 0 you have a
turning point
stationary point
maximum
minium
Thankyou both so much X 
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 17032012 10:02
Yay! Calculus was successfully "explained".

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 17032012 10:19
(Original post by Mr M)
Yay! Calculus was successfully "explained". 
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 17032012 14:12
(Original post by jami74)
Yes I did this. I differentiated to get 0 = ax^2 + bx^2 + c
I then plugged my numbers into a formula b +or square root of b^2  4ac /2a and got two x = numbers.
Then I plugged my x = numbers into my original y= equation and got two sets of coordinates. So far this makes sense to me.
Oh I see! I differentiate and solve for x, which gives me the x part of the coordinate and then use that to find the y part of the coordinate. I feel a bit silly now that I needed that explaining to me
Thankyou both so much X 
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 17032012 21:52
(Original post by Mr M)
Yay! Calculus was successfully "explained".
Am I allowed to carry on asking dumb questions or will I get kicked off the maths board? 
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 17032012 21:55
(Original post by jami74)
But wait! I think there's more...
Am I allowed to carry on asking dumb questions or will I get kicked off the maths board? 
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 17032012 21:58
(Original post by Mr M)
Yay! Calculus was successfully "explained". 
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 17032012 22:57
Okay, does the tangent line always always touch the given point of the curve?
The reason I am asking is because I have a question that asks me to find the equation of the tangent on one point and the equation of the normal on the other point. I differentiated to get y=mx+c and found the m then the y then the c for the first point. Then I drew a graph because I wanted to see what it looked like and sure enough the tangent line touched the right point.
When I did the same for the second point the tangent line runs a tiny bit under the point of the curve. I have done the sums over and over again and can't get different numbers. 
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 17032012 23:01
The tangent touches the curve
Do you want to give the question 
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 17032012 23:02
Ooh can I? Would you mind if I sent it privately because it is an assignment question and I don't want to get into trouble for putting it on the internet?

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 17032012 23:04
Well if it is an assignment problem I would rather not give a solution
If you send the question and solution I will see if it is correct 
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 17032012 23:11
Thankyou. I don't want anyone to tell me the answer, what I really want is to understand it.

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 18032012 21:55
Would anyone mind talking me through how to find the normal please? I know that I have to do 1/m but then what? I've got as far as y= 1/m +c, do I use the same y,m and c numbers that I got when I differentiated it from the original equation?

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 18032012 21:56
(Original post by jami74)
Would anyone mind talking me through how to find the normal please? I know that I have to do 1/m but then what? I've got as far as y= 1/m +c, do I use the same y,m and c numbers that I got when I differentiated it from the original equation? 
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 18032012 22:00
You find the gradient of the tangent, and do 1/that to find the gradient of the normal. Then you just use the gradient of the normal and a point you know is on the line to find c in the y=mx+c in the exact same way you find the equation of a tangent.

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 18032012 22:09
(Original post by Mr M)
You need to multiply that gradient by x. c will not be the same. 
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 18032012 22:54
(Original post by jami74)
Ah that's where I'm going wrong then I need a new c. Yes, of course I do and I realise why now. My x= 1 and my m=1 so does this look right? 1/1(1) = 1. Is that my y= ?
yy1=mt(xx1)
If ur trying to find the normal you replace mt (which is the tangent gradient) with mn (the normal gradient) this is found by doing 1/mt. This is how iw as taught anyways, im not too sure if we're using the same letter to represent the same things.
The y1 is your y coordinate and the x1 is your x coordinate. You move ALL the terms except y to the right hand side of the equals sign and then solve, collecting any like terms as you go. This would give your solution as y=mx+c.
I may be wrong but thats how i was taught the equation of the line, normal or tangent.Last edited by Lovin; 18032012 at 22:55.
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Updated: March 21, 2012
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