Please help explain calculus to me

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    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.
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    (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.
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    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
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    (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
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    Yay! Calculus was successfully "explained".
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    (Original post by Mr M)
    Yay! Calculus was successfully "explained".
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    (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
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    (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?
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    (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....
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    (Original post by Mr M)
    Yay! Calculus was successfully "explained".
    Bah! PRSOM
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    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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    The tangent touches the curve

    Do you want to give the question
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    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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    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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    Thank-you. I don't want anyone to tell me the answer, what I really want is to understand it.
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    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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    (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.
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    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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    (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= ?
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    (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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