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Math Help - what is the derivative of r = t squared/8 + 1 ?

  1. #1
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    what is the derivative of r = t squared/8 + 1 ?

    Hello,
    Among my collection of Maths books, I have found a great AS Maths book which is, at last, really clearly explaining Applications of Differentiation to me.
    But I have just reached a part, in a worked example, that I don’t understand.
    The question involves a basic formula: r = t2/8 + 1, (the "t2" bit means "t squared") which gives you the radius of an ink spot, the number of seconds after it first appears.
    I don’t understand the last bit of the example. They ask you to find “the rate of increase of the radius after 4 seconds”.
    They say you need to differentiate, I understand that. But I don’t understand how the derivative that they suggest can be right, as I thought you ignored numbers, eg, 2, 3, etc when finding the derivative.(I understand by the way, why numbers are ignored, it’s because eg, 8 can be considered as 8xo etc) But they have KEPT THE EIGHT in their derivative, and I don’t understand why.
    Their derivative is:
    Dr/dt = 2t/8 = t/4,
    ( based on the original function of: r = t2/8 + 1,_("t squared over 8, plus 1)

    And so, they say, after 4 seconds the rate of increase is 4/4 = 1 cm per second.
    Is their derivative correct? And if it is, I would be grateful if anyone could explain to me why the 8 in the original equation is not just ignored when calculating the derivative?
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  2. #2
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    Quote Originally Posted by protractor View Post
    Hello,
    Among my collection of Maths books, I have found a great AS Maths book which is, at last, really clearly explaining Applications of Differentiation to me.
    But I have just reached a part, in a worked example, that I donít understand.
    The question involves a basic formula: r = t2/8 + 1, (the "t2" bit means "t squared") which gives you the radius of an ink spot, the number of seconds after it first appears.
    I donít understand the last bit of the example. They ask you to find ďthe rate of increase of the radius after 4 secondsĒ.
    They say you need to differentiate, I understand that. But I donít understand how the derivative that they suggest can be right, as I thought you ignored numbers, eg, 2, 3, etc when finding the derivative.(I understand by the way, why numbers are ignored, itís because eg, 8 can be considered as 8xo etc) But they have KEPT THE EIGHT in their derivative, and I donít understand why.
    Their derivative is:
    Dr/dt = 2t/8 = t/4,
    ( based on the original function of: r = t2/8 + 1,_("t squared over 8, plus 1)

    And so, they say, after 4 seconds the rate of increase is 4/4 = 1 cm per second.
    Is their derivative correct? And if it is, I would be grateful if anyone could explain to me why the 8 in the original equation is not just ignored when calculating the derivative?
    The 8 is not ignored because it is the coefficent of t^2. Only constant terms can be ignored so the 8 cannot be ignored.

    If you imagine differentiation as the gradient of a graph at any given point then it is clear that the graph of \frac{t^2}{8} is not the same as the graph of  t^2

    The book's derivative is correct, keep the 8
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  3. #3
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    Quote Originally Posted by protractor View Post
    Hello,
    Among my collection of Maths books, I have found a great AS Maths book which is, at last, really clearly explaining Applications of Differentiation to me.
    But I have just reached a part, in a worked example, that I donít understand.
    The question involves a basic formula: r = t2/8 + 1, (the "t2" bit means "t squared") which gives you the radius of an ink spot, the number of seconds after it first appears.
    I donít understand the last bit of the example. They ask you to find ďthe rate of increase of the radius after 4 secondsĒ.
    They say you need to differentiate, I understand that. But I donít understand how the derivative that they suggest can be right, as I thought you ignored numbers, eg, 2, 3, etc when finding the derivative.(I understand by the way, why numbers are ignored, itís because eg, 8 can be considered as 8xo etc) But they have KEPT THE EIGHT in their derivative, and I donít understand why.
    Their derivative is:
    Dr/dt = 2t/8 = t/4,
    ( based on the original function of: r = t2/8 + 1,_("t squared over 8, plus 1)

    And so, they say, after 4 seconds the rate of increase is 4/4 = 1 cm per second.
    Is their derivative correct? And if it is, I would be grateful if anyone could explain to me why the 8 in the original equation is not just ignored when calculating the derivative?
    \frac{t^2}{8} = \frac{1}{8}t^2.

    What happens to a constant coefficient when you differentiate?
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  4. #4
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    Quote Originally Posted by Prove It View Post
    \frac{t^2}{8} = \frac{1}{8}t^2.

    What happens to a constant coefficient when you differentiate?
    Nothing "happens" to it.

    You can, if you like, think of it as using the product rule:
    (fg)'= f'g+ fg'. If f is a constant, then the derivative of it is a constant so:
    (fg)'= 0(g)+ fg'= fg'. Only the non-constant function is differentiated but you still have the constant.

    If c is a constant, g any function of x, (cf)'= cf'.
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  5. #5
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    1/8 is a co-efficient of t squared!

    Thanks very much for your great answers e-pi, Prove it, and Halls of Ivy, which I understand perfectly. That is so interesting, the thing about 1/8 being a co-efficient of t squared.It's so helpful to have such useful advice so quickly, it's tiresome when one little thing like that holds you up, thank you very much.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Nothing "happens" to it.

    You can, if you like, think of it as using the product rule:
    (fg)'= f'g+ fg'. If f is a constant, then the derivative of it is a constant so:
    (fg)'= 0(g)+ fg'= fg'. Only the non-constant function is differentiated but you still have the constant.

    If c is a constant, g any function of x, (cf)'= cf'.
    Yes, something does.

    It's multiplied by the exponent, before the exponent has 1 subtracted...
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