Author Topic: TrueTears question thread (Read 68068 times) Tweet Share

TrueTears · « **Reply #450 on:** July 28, 2009, 11:06:58 pm »

Ahhh ok, thanks Mao!

TrueTears · « **Reply #451 on:** July 29, 2009, 03:54:03 pm »

How to change $x^3-21x^2+147x$ into cubic "turning point" form, ie $(x-7)^3 +343$ ?

Thanks!

dcc · « **Reply #452 on:** July 29, 2009, 04:12:20 pm »

Such a form doesn't necessarily exist, however for this 'special case' you can do:
$x^3 - 3a x^2 + 3a^2 x = (x - a)^3 + a^3$ , where $a = 7$ in the above example.

EDIT:

And I must add, I don't really see a purpose to this form for a cubic. Only a subset of the cubics can be expressed in such a way, and it doesn't tell you much about anything at all (quite unlike the turning point form for a parabola, where you can immediately recognise the minimum value of the function and where it occurs).

TrueTears · « **Reply #453 on:** July 30, 2009, 07:05:49 pm »

thanks dcc

TrueTears · « **Reply #454 on:** August 01, 2009, 07:15:20 pm »

Just this question from an exam... I think answer is wrong can anyone show me what they get when they do this question?

Shade the required region of:

$y \le x (\sqrt{3} -2)$ n $y > x(-\sqrt{3} - 2)$ n $x^2+y^2 < \frac{1}{3}$

Many thanks!

Damo17 · « **Reply #455 on:** August 01, 2009, 07:38:44 pm »

Quote from: TrueTears on August 01, 2009, 07:15:20 pm

Just this question from an exam... I think answer is wrong can anyone show me what they get when they do this question?

Shade the required region of:

$y \le x (\sqrt{3} -2)$ n $y > x(-\sqrt{3} - 2)$ n $x^2+y^2 < \frac{1}{3}$

Many thanks!

Haven't done these questions in a long time. I think it is the shaded area between all 3 graphs in the 4th quadrant.

The circle and the line with steep gradient are meant to be dotted.

TrueTears · « **Reply #456 on:** August 01, 2009, 08:24:33 pm »

Yeah I got that as well, ok thanks Damo! Now I know the answers are wrong

TrueTears · « **Reply #457 on:** August 02, 2009, 02:52:54 pm »

I did this question but I dono if my method is acceptable... anyways how would you guys attempt this question?

kamil9876 · « **Reply #458 on:** August 02, 2009, 03:15:17 pm »

(1) $\frac{dx}{dt}=5$ so use chain rule after finding $\frac{dx}{d\theta}$ from implicit differentiation.

(2) After implicitly differentiating and finding $\frac{dx}{d\theta}$ , notice the equation $5t=x$ . Use the identity:

$\frac{dx}{d\theta}=\frac{d(5t)}{d\theta}=5\frac{dt}{d\theta}$ and reciprocalise etc.

method 2 only worked because it's a linear function, more complicated ones work better with the chain rule (method 1)

TrueTears · « **Reply #459 on:** August 02, 2009, 04:42:03 pm »

Quote from: kamil9876 on August 02, 2009, 03:15:17 pm

(1) $\frac{dx}{dt}=5$ so use chain rule after finding $\frac{dx}{d\theta}$ from implicit differentiation.

(2) After implicitly differentiating and finding $\frac{dx}{d\theta}$ , notice the equation $5t=x$ . Use the identity:

$\frac{dx}{d\theta}=\frac{d(5t)}{d\theta}=5\frac{dt}{d\theta}$ and reciprocalise etc.

method 2 only worked because it's a linear function, more complicated ones work better with the chain rule (method 1)

True but that's not using "implicit differentiating" is it?

when you work out $\frac{d\theta}{dx}$ from the $tan(\theta) = \frac{40}{x}$ , that is not implicit differentiation, that is just differentiating with respect to x.

kamil9876 · « **Reply #460 on:** August 02, 2009, 04:50:06 pm »

"Normal Differentiation" is a subset of implicit differentiation

. Differentiating both sides by x is implicit differentiation in this case, after all you have to use the chain rule for both sides and all that shiz that normally comes with implicit diff.

But yes, if you were to rearrange first then it woudl be 'regular' differentiation lol.

$<br />\frac{d}{dx}(tan\theta)=\frac{d}{dx}(40x^{-1})$
$\frac{d\theta}{dx} \frac{d(tan\theta)}{d\theta}=\frac{-40}{x^2}$
$\frac{d\theta}{dx} sec^2 \theta=\frac{-40}{x^2}$
$\frac{d\theta}{dx}=cos^2 \theta \frac{-40}{x^2}$

TrueTears · « **Reply #461 on:** August 02, 2009, 05:13:16 pm »

Thanks kamil!

TrueTears · « **Reply #462 on:** August 02, 2009, 05:21:12 pm »

How about this innocent looking question but somewhere I get wrong answer...

Solve the differential equation $\frac{dy}{dx}= \frac{-y}{2}$ given that when x = 0 y = -1

kamil9876 · « **Reply #463 on:** August 02, 2009, 05:27:40 pm »

$\frac{dx}{dy}=\frac{-2}{y}$
$x=-2 ln|y| +c$

Here's a fun way of getting rid of the modulus sign:

$x=-2 ln Ay$

$0=-2ln(-A)<br />$
$\imples A=-1$

You probably forgot the modulus and thought that the function is +y inside the argument but really it has a - as shown by the negative A value.

TrueTears · « **Reply #464 on:** August 02, 2009, 05:40:19 pm »

lol true true thanks!

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