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1. 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.
2. (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.
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.

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 co-ordinate of y.
3. 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
4. (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.

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 co-ordinate of y.
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 co-ordinates. 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
Oh I see! I differentiate and solve for x, which gives me the x part of the co-ordinate and then use that to find the y part of the co-ordinate. I feel a bit silly now that I needed that explaining to me

Thank-you both so much X
5. Yay! Calculus was successfully "explained".
6. (Original post by Mr M)
Yay! Calculus was successfully "explained".
7. (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 co-ordinates. So far this makes sense to me.

Oh I see! I differentiate and solve for x, which gives me the x part of the co-ordinate and then use that to find the y part of the co-ordinate. I feel a bit silly now that I needed that explaining to me

Thank-you both so much X
8. (Original post by Mr M)
Yay! Calculus was successfully "explained".
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?
9. (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?
They aren't dumb questions....
10. (Original post by Mr M)
Yay! Calculus was successfully "explained".
Bah! PRSOM
11. 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.
12. The tangent touches the curve

Do you want to give the question
13. 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?
14. 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
15. Thank-you. I don't want anyone to tell me the answer, what I really want is to understand it.
16. 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?
17. (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?
You need to multiply that gradient by x. c will not be the same.
18. 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.
19. (Original post by Mr M)
You need to multiply that gradient by x. c will not be the same.
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= ?
20. (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= ?
im not too sure what youv done here. But the equation of the tangents is ---
y-y1=mt(x-x1)

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 co-ordinate and the x1 is your x co-ordinate. 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.

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