Author Topic: Bucket's Questions (Read 57198 times) Tweet Share

Glockmeister · « **Reply #150 on:** March 20, 2008, 09:42:02 pm »

Yep, curves must be 'smooth' and continuous, IRRC.

I'm sure there's a more mathematical way of stating this.

bucket · « **Reply #151 on:** March 20, 2008, 10:35:26 pm »

Quote from: Glockmeister on March 20, 2008, 09:42:02 pm

Yep, curves must be 'smooth' and continuous, IRRC.

I'm sure there's a more mathematical way of stating this.

rofl thats exactly what my tutor told me.

bucket · « **Reply #152 on:** March 20, 2008, 11:45:09 pm »

for the function $3(x-1)^2$ , how do you find when $f'(x)>0$ ?

Collin Li · « **Reply #153 on:** March 20, 2008, 11:46:47 pm »

bucket · « **Reply #154 on:** March 21, 2008, 01:06:41 am »

omg that was such a stupid question rofl.
this one i dont know what to do with either

Given that the curve $y=ax^2+\frac{b}{x}$ has a gradient of -5 at the point (2,-2), find the value of a and b

Collin Li · « **Reply #155 on:** March 21, 2008, 01:24:58 am »

$\frac{dy}{dx} = 2ax - \frac{b}{x^2}$

At the point (2, -2), the gradient is -5. In other words, when $x=2$ , $\frac{dy}{dx} = -5$

$\implies -5 = 2a(2) -\frac{b}{2^2} \implies -5 = 4a - \frac{b}{4}\mbox{ ------ (1)}$

Now, to get another equation (trying to set up simultaneous equations):

The point (2, -2) on the curve tells us that when $x=2$ , $y=-2$

$\implies -2 = a(2)^2 + \frac{b}{2} \implies -2 = 4a + \frac{b}{2}\mbox{ ------ (2)}$

Subtracting (1) from (2) yields:

$-2 + 5 = \frac{b}{2} - \frac{b}{4} \implies 3 = \frac{b}{4} \implies b = 12$

Therefore, $a = -2$

unknown id · « **Reply #156 on:** March 21, 2008, 06:14:14 pm »

Quote from: coblin on March 21, 2008, 01:24:58 am

Subtracting (1) from (2) yields:

$-2 + 5 = \frac{b}{2} - \frac{b}{4} \implies 3 = \frac{b}{4} \implies b = 12$

Therefore, $a = -2$

$-2 + 5 = \frac{b}{2} - \frac{-b}{4}$

$3 = \frac{3b}{4}$

$b = 4$

Thus, $a = -1$

bucket · « **Reply #157 on:** April 12, 2008, 02:14:54 am »

Hey guys, I'm back again

I seem to be able to do these problems half way but then don't know where to go afterwards.

'Find the derivative of $4x^2(2x^2+1)^2$ with respect to x using the product rule.'

I get up to $(4x^2 \cdot8x(2x^2+1)) + ((2x^2+1)^2\cdot8x)$ but then they seem to simplify the answer quite a bit.

The answer they get is $8x(2x^2+1)(6x^2+1)$ , how do they get to this?

Is my working out completely wrong? lol

ed_saifa · « **Reply #158 on:** April 12, 2008, 08:12:17 am »

'Find the derivative of $4x^2(2x^2+1)^2$ with respect to x using the product rule.'

"Let $u= 4x^2 and v= (2x^2+1)^2$

Therefore $\frac{d}{dx} 4x^2(2x^2+1)^2 =v\frac{du}{dx} + u\frac{dv}{dx}$

$\frac{du}{dx}= 8x \frac{dv}{dx}= 8x(2x^2+1)$

Putting this all together gives

$8x(2x^2+1)^2 + 4x^2.2.4x(2x^2+1)$

$8x(2x^2+1)^2 +32x^3(2x^2+1)$

Factor out $8x$

$8x( (2x^2+1)^2 +4x^2(2x^2+1)$

Factor out $(2x^2+1)^2$

$8x(2x^2+1)(2x^2+1+4x^2)$

$8x(2x^2+1)(6x^2+1)$

bucket · « **Reply #159 on:** April 12, 2008, 03:29:26 pm »

mm thanks ed

bucket · « **Reply #160 on:** May 25, 2008, 11:40:05 pm »

I have a questionnn........................................

Find the derivative of $e^{4x+1}(x+1)^2$

and also of $e^{-4x}\sqrt{(x+1)}$

Glockmeister · « **Reply #161 on:** May 26, 2008, 12:32:58 am »

They both use the chain rule

e.g The first one: $u=e^{4x+1}$ $v=(x+1)^{2}$

When you find out the derivative of v, you use the chain rule

$\frac{dv}{dx} = 2(x+1)$

And then you apply for the formula.

It's a similar technique for the second one.

If you can't solve it, I'll work out the solution tomorrow, when I'm a bit more awake.

ell · « **Reply #162 on:** May 26, 2008, 01:06:06 pm »

For the second one:
$u=e^{-4x}$ and $v=\sqrt{x+1}=(x+1)^{1/2}$

$\frac{du}{dx}=-4e^{-4x}$ and $\frac{dv}{dx}=\frac{1}{2\sqrt{x+1}}$ (chain rule)

Apply product rule:
$\begin{align*} \frac{dy}{dx}&=v\frac{du}{dx}+u\frac{dv}{dx}\\ &=\sqrt{x+1}\cdot-4e^{-4x}+e^{-4x}\cdot\frac{1}{2\sqrt{x+1}}\\ &=\frac{-4\sqrt{x+1}}{e^{4x}}+\frac{1}{2e^{4x}\sqrt{x+1}}\\\end{align*}$

Get a common denominator:
$\begin{align*} \frac{dy}{dx}&=\frac{-4\sqrt{x+1}}{e^{4x}}\cdot\frac{2\sqrt{x+1}}{2\sqrt{x+1}}+\frac{1}{2e^{4x}\sqrt{x+1}}\\ &=\frac{-8(x+1)+1}{2e^{4x}\sqrt{x+1}}\\ &=\frac{-8x-7}{2e^{4x}\sqrt{x+1}}\\ \end{align*}$

bucket · « **Reply #163 on:** May 26, 2008, 10:51:09 pm »

ahh thanks alot both of you.
x

bucket · « **Reply #164 on:** June 17, 2008, 09:33:21 pm »

I have forgotten how to derive after two weeks w/o maths :S.

$f(x)=\dfrac{1}{8}(x-1)^3(8-3x)+1$ , show that $f'(x)=\dfrac{3}{8}(x-1)^2(9-4x)$ and specify when $f'(x)\geq0$

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