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    Find an equation of the normal to the curve y=e^3x -2 at the point (1,e)

    and would someone be so kind as to show me their working too?
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    (Original post by Zarantulas)
    Find an equation of the normal to the curve y=e^3x -2 at the point (1,e)

    and would someone be so kind as to show me their working too?
    What are your thoughts on this question?

    Spoiler:
    Show

    dy/dx= 3e^3x
    y= -1/m
    y=1/3e^3x +c

    y-e/x-1 = 1/3e^3x
    y-e=3e^-3x (x-1)
    y=3e^-3x (x-1)+e


    Answers if you really can't do it^^
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    (Original post by Zarantulas)
    Find an equation of the normal to the curve y=e^3x -2 at the point (1,e)

    and would someone be so kind as to show me their working too?
    Eh doing a question in its entirety is pretty boring after getting 100 UMS in C3 so I'll walk you through it.

    Step 1: Differentiate the curve.

    Step 2: Find the gradient of the tangent at the point (1,e) by substituting the appropriate value into the differential.

    Step 3: You should know what to do to get the normal gradient of a given gradient. So do that.

    Step 4: Use y-b=m(x-a) to construct your equation.
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    (Original post by RDKGames)
    Eh doing a question in its entirety is pretty boring after getting 100 UMS in C3 so I'll walk you through it.

    Step 1: Differentiate the curve.

    Step 2: Find the gradient of the tangent at the point (1,e) by substituting the appropriate value into the differential.

    Step 3: You should know what to do to get the normal gradient of a given gradient. So do that.

    Step 4: Use y-b=m(x-a) to construct your equation.
    I can tell you're popular
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    (Original post by Zarantulas)
    Find an equation of the normal to the curve y=e^3x -2 at the point (1,e)

    and would someone be so kind as to show me their working too?
    I'm not going to answer it but ill guide you through it lol.

    First, differentiate to get the gradient function. Remember when you differentiate e^x you get e^x and you multiply whatever the derivative of the x component by the coefficient of e. (Essentially the derivative of e^g(x) is g'(x)e^g(x) )

    Then, sub in your x value to get the gradient at that point. The question asks for the normal, so your gradient will be the negative reciprocal of what you've just worked out.

    Finally sub your gradient value and the point you're given into the equation of a line and hey presto there's your answer.
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    (Original post by okey)
    I can tell you're popular
    Indeed I am :holmes:

    Next time they should attempt the question before asking someone else to do it.
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    How do you know i haven't attempted the question?
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    (Original post by Zarantulas)
    How do you know i haven't attempted the question?
    Because you haven't shown any attempt on it and for all we know you could be lying lol.

    For the sake of others, it is much easier for us to just go through your attempt and point out where you go wrong as opposed to us doing the entire question for you.
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    Surprised no one has yet pointed out that the point (1,e) does not actually lie on the given curve.
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    If it pleases you, this is my attempt:

    y =e^3x-2

    dy/dx = 3e^3x -2

    sub in x = 1

    3e

    gradient of tangent = 3e so gradient of tangent = -1/3e

    y = mx + c
    y = -1/3e x + c
    x =1 and y = e

    e = -1/3 e + c

    e + 1/3 e = c

    and I am not sure where to go from here







    (Original post by RDKGames)
    Because you haven't shown any attempt on it and for all we know you could be lying lol.

    For the sake of others, it is much easier for us to just go through your attempt and point out where you go wrong as opposed to us doing the entire question for you.
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    (Original post by Zarantulas)
    If it pleases you, this is my attempt:

    y =e^3x-2

    dy/dx = 3e^3x -2
    ...
    Constant disappears when differentating.
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    so would it come to e + 1/3 e^3 = c to where i got up to?
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    (Original post by RDKGames)
    Eh doing a question in its entirety is pretty boring after getting 100 UMS in C3 so I'll walk you through it.
    Was this comment absolutely necessary?
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    (Original post by Ano9901whichone)
    Was this comment absolutely necessary?
    No. But then again a question without an attempt is not necessary either.
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    (Original post by RDKGames)
    Eh doing a question in its entirety is pretty boring after getting 100 UMS in C3 so I'll walk you through it.

    Step 1: Differentiate the curve.

    Step 2: Find the gradient of the tangent at the point (1,e) by substituting the appropriate value into the differential.

    Step 3: You should know what to do to get the normal gradient of a given gradient. So do that.

    Step 4: Use y-b=m(x-a) to construct your equation.
    this^
 
 
 
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