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

1. ## 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?

2. Originally Posted by protractor
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?
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

3. Originally Posted by protractor
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?
$\frac{t^2}{8} = \frac{1}{8}t^2$.

What happens to a constant coefficient when you differentiate?

4. Originally Posted by Prove It
$\frac{t^2}{8} = \frac{1}{8}t^2$.

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'.

5. ## 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.

6. Originally Posted by HallsofIvy
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...