[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.

[Lovin]

(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.
That'll give you your co-ordinates

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.

[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

[jami74]

(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 co-ordinates

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

[Mr M]

Yay! Calculus was successfully "explained".

[TenOfThem]

(Original post by Mr M)
Yay! Calculus was successfully "explained".

[Lovin]

(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

Your welcome sweetie

[jami74]

(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?

[bananarama2]

(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....

[shamika]

(Original post by Mr M)
Yay! Calculus was successfully "explained".

Bah! PRSOM

[jami74]

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.

[TenOfThem]

The tangent touches the curve

Do you want to give the question

[jami74]

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?

[TenOfThem]

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

[jami74]

Thank-you. I don't want anyone to tell me the answer, what I really want is to understand it.

[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?

[Mr M]

(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.

[Dj.Clay]

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.

[jami74]

(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= ?

[Lovin]

(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.