Author Topic: TrueTears question thread (Read 66216 times) Tweet Share

/0 · « **Reply #60 on:** December 30, 2008, 04:55:59 pm »

Quote from: TrueTears on December 30, 2008, 04:36:04 pm

ah thanks.

also, show that the parabola with equation $ax^2+bx+c$ has no points of inflexion.

$\frac{dy}{dx} = 2ax+b$

$\frac{d^2y}{dx^2} = 2a$

Assuming $a \neq 0$ , $\frac{d^2y}{dx^2} \neq 0$ , so the graph will always be either concave up or concave down.

humph · « **Reply #61 on:** December 30, 2008, 09:28:44 pm »

Quote from: TrueTears on December 30, 2008, 04:36:04 pm

ah thanks.

also,

1. show that the parabola with equation $ax^2+bx+c$ has no points of inflexion.

2. Determine the points of inflexion for the following function:

$(x^2-1)(x^2+1) = y$

$\implies \frac{dy}{dx} = 4x^3$

$\implies \frac{d^2y}{dx^2} = 12x^2$

so point of inflexion occurs when $\frac{d^2y}{dx^2} = 0$ $\implies x = 0$ . But this is not a point of inflexion, why is that?

If $x$ is a point of inflection of $f$ , then $f''(x)=0$ . Unfortunately, the converse is not true: $f''(x)=0$ does not necessarily imply that $x$ is a point of inflection of $f$ . That is, $f''(x)=0$ is necessary but not sufficient for $x$ to be a point of inflection of $f$ .

TrueTears · « **Reply #62 on:** December 30, 2008, 10:07:18 pm »

ah i see, so what are the conditions for point of inflexion?

And also this question:

The equation of a curve C is $x^3+xy+2y^3 = k$ , where $k$ is a constant

a) find $\frac{dy}{dx}$ $\implies$ $\frac{dy}{dx} = \frac{-3x^2-y}{x+6y^2}$

C does have a tangent parallel to the y axis

b) show that the y coordinate at the point of contact satisfies $216y^6+4y^3+k = 0$

how do u do part b)?

many thanks

humph · « **Reply #63 on:** December 31, 2008, 01:45:05 am »

Quote from: TrueTears on December 30, 2008, 10:07:18 pm

ah i see, so what are the conditions for point of inflexion?

I think it involves calculating $f''$ at points near and either side of the point you're interested in. Can't remember the exact details though, but it should be in a textbook.

Quote from: TrueTears on December 30, 2008, 10:07:18 pm

And also this question:

The equation of a curve C is $x^3+xy+2y^3 = k$ , where $k$ is a constant

a) find $\frac{dy}{dx}$ $\implies$ $\frac{dy}{dx} = \frac{-3x^2-y}{x+6y^2}$

C does have a tangent parallel to the y axis

b) show that the y coordinate at the point of contact satisfies $216y^6+4y^3+k = 0$

how do u do part b)?

many thanks

" $C$ does have a tangent parallel to the $y$ -axis" means that the gradient is infinite at this point (this is quite dodgy mathematically but hopefully you understand what I mean geometrically - a straight vertical line is kinda like a linear function with infinite gradient).
Anyway, this will obviously happen when the denominator of $f'$ is equal to zero. You found this to be $x+6y^2$ , so we have that $x+6y^2=0 \implies x = -6y^2$ .
So what we have is that when the tangent of the curve is parallel to the $y$ -axis, and so the gradient is "infinite", we must have that the point $(x,y)$ at which this occurs must satisfy the relation $x = -6y^2$ . In particular, we can substitute this value of $x$ into the equation of the curve (as this point lies on the curve) in order to find that $216y^6+4y^3+k = 0$ .

TrueTears · « **Reply #64 on:** December 31, 2008, 01:50:17 am »

ah yessss thanks so much , that was wat i was confused about, coz a vertically has undefined gradient lol

TrueTears · « **Reply #65 on:** December 31, 2008, 02:47:39 pm »

yeah and just back to the point of inflexion stuff, my book says "A point of inflexion will occur at $x = \alpha$ if $f''(x) = 0$ , and $f''(\alpha + \beta)$ and $f''(\alpha - \beta)$ have different signs."

So to work out the points of inflexion for $f(x) = sin(x)$ for $x \in [0,2\pi]$

$f''(x) = -sin(x)$

$\implies -sin(x) = 0$

$\implies x = 0 , \pi , 2\pi$

so $f''(0+1) = -0.841$ ... and $f''(0-1) = 0.841$ ... So there will be a point of inflexion at $x=0$ since both signs are different.

But this is where i don't get, if you do the same to $\pi, 2\pi$ you get:

$f''(\pi+1) = 0.841$ ... and $f''(\pi-1) = -0.841$ ... So both signs are different, so there should be a point of inflexion at $x = \pi$ . However my book's answers says there isn't. The same goes for $2\pi$ . Can someone clarify this?

Matt The Rat · « **Reply #66 on:** December 31, 2008, 03:10:02 pm »

For Spec, I don't think you need to be too worried expect about points of inflexion. Just remember $f"(x)=0$ and you should be right. If not, blame Coblin; it's what he told me. But the way the book describes it is correct.

/0 · « **Reply #67 on:** December 31, 2008, 04:36:32 pm »

I think $x=0,\pi,2\pi$ should all be points of inflection. An easy way to visualise it is as if you are driving a car on the curve. Every time you are not 'turning' left or right, but driving straight (momentarily) you will get a point of inflection.

TrueTears · « **Reply #68 on:** December 31, 2008, 06:42:01 pm »

Quote from: /0 on December 31, 2008, 04:36:32 pm

I think $x=0,\pi,2\pi$ should all be points of inflection. An easy way to visualise it is as if you are driving a car on the curve. Every time you are not 'turning' left or right, but driving straight (momentarily) you will get a point of inflection.

ah, kk so i guess the book's answer is wrong o.o

And also some other questions

1. Consider $f(x) = tan^{-1} + tan^{-1}(\frac{1}{x})$ , $x$ does not equal 0

If x > 0, find $f(x)$ .

My book does $f(x) = tan^{-1}(x) + \frac{\pi}{2} - tan^{-1}(x)$ . How do u get the $\frac{\pi}{2}$ ?

2. $A = \frac{\sqrt{4+p^2}}{2} + \frac{p^2}{2\sqrt{4+p^2}} -1$ . Show that $A \ge 0$ for all $p$

3. Co-ordinate Z is described by $x = 3( cos(\theta) - sin(\theta) )$ and $y = 2( sin(\theta) + cos(\theta) )$ , find the locus of Z as $\theta$ varies.

Many thanks again, i really appreciate all your helps XD

/0 · « **Reply #69 on:** January 01, 2009, 02:29:51 am »

1.
The derivative of the entire expression is 0, so the graph is constant.

$f(1) = \frac{\pi}{2}$

Alternatively, derive from a triangle... or use the fact that $\tan{f(x)}= undefined$ to deduce that $f(x) = \frac{\pi}{2}$ .

2.

$A = \frac{p^2+2}{\sqrt{p^2+4}}-1$

$\Rightarrow \frac{dA}{dp} = \frac{p(p^2+6)}{(p^2+4)^\frac{3}{2}}$

If the graph has non-negative gradient, then:

$\frac{dA}{dp} \geq 0 \Rightarrow p(p^2+6) \geq 0$

Since $p(p^2+6)$ is the product of two numbers, and since $p^2+6$ is always greater than 0, it follows that $p \geq 0$ .

So the graph is rising for $p \geq 0$ .

When $p=0$ , $A = 0$

The gradient of the graph is negative for $x<0$ , but the lowest point is A = 0, from which point it starts rising. $\therefore \forall p\in R,\ A \geq 0$ (for all p in R)

3.

$\frac{x}{3} = \cos{\theta}-\sin{\theta}$ , $\frac{y}{2} = \sin{\theta}+\cos{\theta}$

$\Rightarrow \left(\frac{x}{3}\right)^2+\left(\frac{y}{2}\right)^2=2$

$\frac{x^2}{9}+\frac{y^2}{4}=2$

TrueTears · « **Reply #70 on:** January 01, 2009, 08:46:45 pm »

hm i get Q 2 and Q 3, but still cloudy on Q 1 o.O

and also just this Q

Find an anti derivative to $cot^2(x)$

ell · « **Reply #71 on:** January 01, 2009, 09:34:54 pm »

Quote from: TrueTears on January 01, 2009, 08:46:45 pm

hm i get Q 2 and Q 3, but still cloudy on Q 1 o.O

and also just this Q

Find an anti derivative to $cot^2(x)$

$\int \cot^2(x)\ dx$

$=\int \tan^2\left(\frac{\pi}{2}-x\right)\ dx$

$=\int \sec^2\left(\frac{\pi}{2}-x\right)-1\ dx$

$=-\tan \left(\frac{\pi}{2}-x\right) - x +C$

$=-\cot (x) - x +C$

TrueTears · « **Reply #72 on:** January 02, 2009, 03:02:05 pm »

thanks ell,

also

1. $\int \frac{e^{2x}}{1+e^x} dx$ just a bit stuck lol

2. $\int sin(3x)cos^5(3x) dx$ i did this question, but my answer looks totally different from book, wondering if someone can confirm if it is right or not.

so $\int sin(3x)cos^5(3x) dx$ = $\int cos^4(3x)cos(3x)sin(3x) dx$

$\implies \int (1-sin^2(3x))^2cos(3x)sin(3x) dx$

let $u = sin(3x)$

$\implies \frac{du}{dx} = 3cos(3x)$

$\implies du = 3cos(3x)dx$

$\implies \frac{1}{3} \int 3(1-sin^2(3x))^2cos(3x)sin(3x) dx$

$\implies \frac{1}{3} \int (1-u^2)^2u du$

$\implies \frac{1}{3} \int u^5 - 2u^3 + u du$

$\implies \frac{1}{3} (\frac{1}{6}u^6 - \frac{1}{2}u^4 + \frac{1}{2}u^2)$

$\implies \frac{1}{18}u^6 - \frac{1}{6}u^4 + \frac{1}{6}u^2$

sub $u = sin(3x)$ back in and we get $\frac{1}{18}sin(3x)^6 - \frac{1}{6}sin(3x)^4 + \frac{1}{6}sin(3x)^2 + c$

However answer in my book has $-\frac{1}{18}cos^6(3x) + c$

Is that a different form, or is my answer wrong?

ed_saifa · « **Reply #73 on:** January 02, 2009, 04:58:54 pm »

$\int \frac{e^{2x}}{1+e^x} dx$

$\int \frac{e^{x}.e^{x}}{1+e^x} dx$

Let $u= 1 + e^x$

Rearranging gives $e^x= u-1$

$\frac{du}{dx}=e^x$

$dx=\frac{du}{e^x}$

$\int \frac{e^{2x}}{1+e^x} dx$ = $\int \frac{(u-1)^2}{u}.\frac{du}{u-1}$

= $\int \frac{u-1}{u}du$

= $\int 1-\frac{1}{u}du$

= $u- \ln|u| + C$

= $e^x + 1 -ln|e^x +1| + C$

For question 2 you are correct.

TrueTears · « **Reply #74 on:** January 02, 2009, 05:03:54 pm »

ahhh thank you very much ed_saifa

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