# Derivatives and other fancy stuff

• Apr 20th 2013, 10:12 PM
moob
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
• Apr 20th 2013, 10:17 PM
MarkFL
Re: Derivatives and other fancy stuff
Computing a derivative by first principles means using the definition:

$f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
• Apr 20th 2013, 11:25 PM
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.