Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

Please help explain calculus to me

Announcements Posted on
Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
Win top prizes in our mega results day giveaway... 24-07-2015
  1. Offline

    ReputationRep:
    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. Offline

    ReputationRep:
    (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.
  3. Offline

    ReputationRep:
    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. Offline

    ReputationRep:
    (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
  5. Offline

    ReputationRep:
    Yay! Calculus was successfully "explained".
  6. Offline

    ReputationRep:
    (Original post by Mr M)
    Yay! Calculus was successfully "explained".
  7. Offline

    ReputationRep:
    (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
  8. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (Original post by Mr M)
    Yay! Calculus was successfully "explained".
    Bah! PRSOM
  11. Offline

    ReputationRep:
    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. Offline

    ReputationRep:
    The tangent touches the curve

    Do you want to give the question
  13. Offline

    ReputationRep:
    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. Offline

    ReputationRep:
    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. Offline

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

    ReputationRep:
    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. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    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. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (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.

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: March 20, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Do you think you'll do better or worse than your predicted grades?
New on TSR

Results are coming...

No sweat. Here's all you need to make sure you're ready

Study resources
x

Think you'll be in clearing or adjustment?

Hear direct from unis that want to talk to you

Get email alerts for university course places that match your subjects and grades. Just let us know what you're studying.

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.