# Math Help - Derivatives and other fancy stuff

1. ## Derivatives and other fancy stuff

Having trouble with a couple of questions

The first one:
The height of a projectile at a particular time (measured in seconds) is given by h(t)=10t(1-t)m. Find the rate of change of height from first principles (using the definition of the rate of change).

- How do I find the rate of change of height and I'm a bit confused on what 'first principles' are.

The second: Consider the function f(x)=x(4-x). Find the equation of the tangent to the graph of f(x) at x=2. [NOTE: when calculating f'(2), use first principles.]

- Again I'm not too sure on the 'first principles' thing.

Any help is much appreciated,

moob

2. ## Re: Derivatives and other fancy stuff

Computing a derivative by first principles means using the definition:

$f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

3. ## Re: Derivatives and other fancy stuff

First one:
$h(t)=10t(1-t)$

$\Rightarrow h(t)= 10t-10t^2$

$h'(t) = \lim_{\delta\rightarrow 0} \frac{h(t+\delta)-h(t)}{\delta}$

I've used delta so as not to confuse the function with the limiting value.

$h'(t) = \lim_{\delta\rightarrow 0} \frac{10(t+\delta)-10(t+\delta)^2-(10t-10t^2)}{\delta}$

You just need to expand some terms.. cancel like terms, factor out a 'h' from the numerator and cancel out the bottom 'h', take the limit and you're done.

Second one:

$\Rightarrow f(x)=4x-x^2$

Same deal with first principles, this time we have an 'x' value.

$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f'(2) = \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$f'(2) = \lim_{h\rightarrow 0} \frac{4(2+h)-(2+h)^2-(4(2)-(2)^2)}{h}$

Same deal as above..
You just need to expand some terms.. cancel like terms, factor out a 'h' from the numerator and cancel out the bottom 'h', take the limit and you're done.

Once we have that equation, that solution with be your gradient (m).
You then use the point gradient formula:

$y - y_1 = m(x - x_1)$

Where
$x_1 = 2$
$y_1 = f(x)= 4(2)-(2)^2$
m = derivative at 2.

Rearrange to get it into the form y = mx + b.