geometric applications of calculus (1 Viewer)

jnney · Aug 13, 2011

sketch

y=x-2/x^2

SpiralFlex · Aug 13, 2011

$y=\frac{x-2}{x^2}$

We can see that x cannot be zero. Keep this in mind.

So $x=0$ is our vertical asymptote.

As $x\rightarrow 0^-$ [As approaches zero from the left.]

What's a number to the left of zero on the $x$ - axis? -0.1. Substitute that in. [If it is positive it will approach positive infinity and if it is negative it will approach negative infinity.]

$y\rightarrow -\infty$

As $x\rightarrow 0^+$ [As approaches zero from the right.]

What's a number to the right of zero on the $x$ ? 0.1. Substitute that in. [If it is positive it will approach positive infinity and if it is negative it will approach negative infinity.]

$y\rightarrow -\infty$

Your graph should now look like this:

http://www.mediafire.com/?sgt7c6v4ll8d28r

Horizontal asymptote:

As $x\rightarrow \infty$ [As we travel along the right of the x axis.]

$y\rightarrow 0^+$ [The graph is a bit above 0.]

As $x\rightarrow -\infty$ [As we travel along the left of the x axis.]

As $y\rightarrow 0^-$ [The graph is a bit below 0.]

Hence, $y=0$ is our horizontal asymptote.

Your graph should look this this:

http://www.mediafire.com/?sgt7c6v4ll8d28r

But there's something interesting about this graph. It's very naughty.

$y=\frac{x-2}{x^2}$

$\frac{dy}{dx}=\frac{1*x^2-(x-2)*2x}{x^4}$

$=\frac{4-x}{x^3}$

For stationary points,

$\frac{dy}{dx}=0$

$4-x=0$

$x=4$

$y=\frac{1}{8}$

So $(4, \frac{1}{8})$

Test for max/min:

$\frac{d^2y}{dx^2}=\frac{2(x-6)}{x^4}$

At $x=4$

$\frac{d^2y}{dx^2}=2(4-6)<0$

Hence $(4, \frac{1}{8})$ is a maximum turning point.

Connect your points

And finally, your graph should look like this:

http://www.mediafire.com/?4x33xdflfgb27bp

jnney · Aug 13, 2011

Oh, so we don't need stationary points etc.?

Alkanes · Aug 13, 2011

It's more like a functions questions.

Shadowdude · Aug 13, 2011

You mean:

$y=\frac{x-2}{x^{2}}$

You can do what SF did, or WolframAlpha if you're really stuck =P

SpiralFlex · Aug 13, 2011

jnney said:
Oh, so we don't need stationary points etc.?

Yes, it's geometrical applications.

Alkanes said:
It's more like a functions questions.

No, it deals with the stationary points. When the graph cuts the horizontal asymptote and approaches infinity.

jnney · Aug 13, 2011

Oh I get it!

How about, 'For all x:y>1, y'<0, and y''>0'

Is it negative exponential?

SpiralFlex · Aug 13, 2011

jnney said:
Oh I get it!

How about, 'For all x:y>1, y'<0, and y''>0'

Is it negative exponential?

I tried to explain it as easily as I can. Many of my classmates struggling with this. Pardon? I don't understand what you are asking.

jnney · Aug 13, 2011

It asks to sketch a possible curve that satisfies that.

Shadowdude · Aug 13, 2011

jnney said:
Oh I get it!

How about, 'For all x:y>1, y'<0, and y''>0'

Is it negative exponential?

e^(-x) goes to 0 as x goes to infinity.

It's probably 1 + e^(-x).

Annum · Feb 1, 2012

Can someone please please please help me with this pleaase

A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of soft drink. Given that 375 mL has a volume of 375cm^3

a. show that the surface area of a can is given by s= 2pier^2 + 750/r

b. find the radius of the can that gives the minimum surface area.

SpiralFlex · Feb 1, 2012

Wow you brought up my post haha.

First of all ALWAYS draw a pretty diagram.

Let the radius of the base of the cylinder (ie the little small circle on the bottom have a radius of r centimetres.

I will also let the height be h.

So the volume is

$V=\pi r^2 h$

Now if I destroy my special can (nnoooooooooooooooooooooo), I get a net shape like this.

And the total surface area is: The area of 2 circles + the rectangle.

That is,

$S=2\pi r^2 + 2\pi r h$

$\therefore=2\pi r(r+h)$

We have now formed two equations.

$375=\pi r^2 h....(1)$

$S=2\pi r(r+h)....(2)$

We need to eliminate a variable in order to solve these equations. I know that h looks like the odd one out, and I don't want to eliminate r because our equation we are aiming to prove contains r. It would be sensible to eliminate h!

From (1),

$h=\frac{375}{\pi r^2}$

Substitute our new result into equation 2.

$S=2\pi r(r+\frac{375}{\pi r^2})$

Some simply multiplication!

$S=2\pi r^2 + \frac{750}{r}$

YAY first part done.

This is a minimum and maximum problem, I have just constructed the equation necessary to begin solve the problem. Let's see what we can do.

I will differentiate the equation,

$\frac{dS}{dr}=4\pi r -\frac{750}{r^2}$

We know for stationary points, (min, max) $\frac{dS}{dr}=0$

$\therefore 4\pi r -\frac{750}{r^2}=0$

$\frac{750}{r^2}=4\pi r$

$4 \pi r^3=750$

Edit: Carrotsticks has finish the problem quicker than I.

Carrotsticks · Feb 1, 2012

Annum said:
Can someone please please please help me with this pleaase

A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of soft drink. Given that 375 mL has a volume of 375cm^3

a. show that the surface area of a can is given by s= 2pier^2 + 750/r

b. find the radius of the can that gives the minimum surface area.

$\\ $Assuming the can is closed and perfectly cylindrical, the outside surface area consists of two circles and a rectangle (from 'unfolding; the curved surface). We will call the radius $ r $ and the height of the can $ h $ ,since they are variable.$ \\\\ $Area Circle $ = \pi r^{2} \\\\ $But we have two of them, so the area of the two circles is $ 2 \pi r^{2} \\\\ $We will now work on the rectangle. We notice that the length of the rectangle is the circumference of the circle, and the height of the rectangle is the same as the height of the circle.$ \\\\ $Hence, the area of the rectangle is Circumference $ \times $ Height $ = 2 \pi r \times h \\\\ $Therefore, the total surface area of the can is $2 \pi r h + 2 \pi r^{2}$

$\\ $But we now have a problem. The expression in the answer is purely in terms of $ r $ whereas our answer is in terms of $ r $ and $ h . \\\\ $This means that we need to somehow find an expression for $ h $ in terms of $ r $, so then we can substitute it back into the formula we had. We will do so by using the fact that the volume of the can is 375mL.$ \\\\ $Lets now find an expression for the volume of the can:$ \\\\ V = \pi r^{2} h = 375 \\\\ \therefore h = \frac{375}{\pi r^{2}} \\\\ $Substituting this back into our original expression yields:$ \\\\ \begin{align*} SA &= 2 \pi r h + 2 \pi r^{2} \\\\ &= 2 \pi r \times \frac{375}{\pi r^{2}} + 2 \pi r^{2} \\\\ &= \frac{750}{r} + 2 \pi r^{2} \end{align*} $ as required.$$

$\\ $We will now try to minimise this. To do this, we will differentiate the function, and then let it be equal to 0. This will give us max/min surface area. For simplicity, denote the surface area of the can by A.$ \\\\ \frac{dA}{dr} = -\frac{750}{r^{2}} + 4 \pi r = 0 \\\\ $Multiply both sides by $ r^{2} $ to acquire: $ \\\\ 4 \pi r^{3} - 750 = 0 \\\\ $Solve for $ r: \\\\ r^{3} = \frac{750}{4 \pi} = \frac{375}{2 \pi} \\\\ \therefore r= \sqrt[3]{\frac{375}{2 \pi}} \doteqdot 3.908 \\\\ $Now we have to check whether it's a maximum or minimum. You can do this using either second derivative test, or table method. However, you can use a bit of logic and deduce that the maximum surface area is infinity, and this occurs when the can is like an infinitely thin pancake. Thus, the expression that we have will give us the minimum surface area.$ \\\\ $Therefore to conclude the question, a radius of... $ \\\\ r= \sqrt[3]{\frac{375}{2 \pi}} \\\\ $... will give us minimum surface area.$$

SpiralFlex · Feb 1, 2012

Ah faster than I.

- Back to English then.

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geometric applications of calculus (1 Viewer)

jnney

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