Intergration calculating centroid of an area

TheBody · 27-09-2015 05:15PM

I think you forgot to attach the image. I'm not seeeing anything.

Hedgecutter · 27-09-2015 05:22PM

Bad wifi.

TheBody · 27-09-2015 05:27PM

You have a mistake in the second line!

[latex]x=8t+\frac{1}{2t^2}=8t+\frac{1}{2}t^{-2} [/latex].

You made the [latex]\frac{1}{2}[/latex] into a 2, which is incorrect.

TheBody · 27-09-2015 05:28PM

I just realised that I missed that earlier today. My bad!!:o

Hedgecutter · 27-09-2015 05:31PM

1/2t^2 does that not become 2t^-2?

TheBody · 27-09-2015 05:32PM

Hedgecutter wrote: »

1/2t^2 does that not become 2t^-2?

No. You are applying the rules of indices to the t bit. The half just comes along for the spin.

Hedgecutter · 27-09-2015 05:41PM

TheBody wrote: »

No. You are applying the rules of indices to the t bit. The half just comes along for the spin.

Ill give it another go.

Hedgecutter · 27-09-2015 05:52PM

Third time lucky

TheBody · 27-09-2015 06:21PM

Almost!! You made a small mistake with the differentiation.

If [latex]x=8t-\frac{1}{2}t^{-2} [/latex] then:

[latex]\frac{dx}{dt}=8+t^{-3} [/latex]. Not the + sign where you have a - sign.

Hedgecutter · 27-09-2015 06:24PM

TheBody wrote: »

Almost!! You made a small mistake with the differentiation.

If [latex]x=8t-\frac{1}{2}t^{-2} [/latex] then:

[latex]\frac{dx}{dt}=8+t^{-3} [/latex]. Not the + sign where you have a - sign.

Perfect. Two more question and turning points finished.

TheBody · 27-09-2015 06:27PM

Well done. Keep up the hard work.

Hedgecutter · 27-09-2015 07:44PM

Wondering if I'm on the right track. Following examples in a book so not sure if my method is right.

Hedgecutter · 27-09-2015 07:45PM

Here's the question

Hedgecutter · 04-10-2015 08:41PM

Stuck here with this implicit function can't remember how to find y

TheBody · 05-10-2015 08:40AM

If x=2, you could always run that value throught the original formula of the ellipse to find the corresponding value of y.

Your differentiating is a bit dodgy too. When implicitly differentiating, we don't put a [latex]\frac{dy}{dx} [/latex] at the start. We would have:

[latex]\frac{x}{2}+\frac{2y}{9}\frac{dy}{dx}=0 [/latex].

See if you can finish it now.

Hedgecutter · 15-10-2015 07:18PM

Struggling with these questions. Not sure how to start them.

dlouth15 · 15-10-2015 11:05PM

Hedgecutter wrote: »

Struggling with these questions. Not sure how to start them.

For thee second question, write down these two formulas:

a. The area [latex]A[/latex] of a cylinder with radius [latex]r[/latex] and height [latex]d[/latex].

b. The volume [latex]V[/latex] of a cylinder with radius [latex]r[/latex] and height [latex]d[/latex].

You are given the total surface area [latex]A[/latex] in equation a. Rearrange this equation so that the height [latex]h[/latex] is the subject of the equation, i.e. you have [latex]h=...[/latex] and you now have an expression for [latex]h[/latex] in terms of [latex]A[/latex] and [latex]r[/latex].

Now substitute this expression for [latex]h[/latex] into equation b.

You now have an expression for volume [latex]V[/latex] in terms of [latex]h[/latex] and [latex]A[/latex]. Call this equation c.

[latex]A[/latex] is fixed and only [latex]h[/latex] varies.

What you have to do now is find the value of [latex]h[/latex] for which [latex]V[/latex] is maximized. This is done by differentiating [latex]V[/latex] with respect to [latex]h[/latex]. You can differentiate again to ensure it is a maximum.

Finally insert the value you obtained for [latex]h[/latex] into equation c. to get your required volume.

Hedgecutter · 16-10-2015 09:43AM

Still struggling to be honest, but had a go at it.

dlouth15 · 16-10-2015 11:09PM

Hedgecutter wrote: »

Still struggling to be honest, but had a go at it.

I'm getting some sort of server error when I click on the link. It might be something to do with work being done on the boards.ie site.

Hedgecutter · 17-10-2015 10:35AM

dlouth15 wrote: »

I'm getting some sort of server error when I click on the link. It might be something to do with work being done on the boards.ie site.

Ya I had trouble last night. Ill post later my solutions for both questions. Answers seem good but I'm hoping my method is sound.

Hedgecutter · 17-10-2015 04:36PM

My solutions

Hedgecutter · 17-10-2015 04:37PM

next question

dlouth15 · 18-10-2015 12:37AM

Hedgecutter wrote: »

next question

This one looks OK.

dlouth15 · 18-10-2015 01:39AM

Here's my solution for the second problem:

Surface area is given by:

[latex]\displaystyle{A=2\pi rh+2\pi r^{2}}[/latex] (Equation a.)

Volume is given by:

[latex] \displaystyle{V=\pi r^{2}h}[/latex] (Equation b.)

Extract [latex]h[/latex] from equation A to give:

[latex] \displaystyle{h=\frac{A}{2\pi r}-r}[/latex] (Equation c.)

Substitute the RHS into equation b. for h.

[latex] \displaystyle{V=\pi r^{2}\left(\frac{A}{2\pi r}-r\right)=\frac{1}{2}Ar-\pi r^{3} } [/latex]

We now have an expression for volume purely in terms of [latex]r[/latex]. We need to maximize this value. Differentiate [latex]V[/latex] with respect to [latex]r[/latex]. (I incorrectly had the differentiation with respect to [latex]h[/latex] in my earlier instructions).

[latex] \displaystyle{ \frac{dV}{dr}=\frac{1}{2}A-3\pi r^{2}} [/latex]

Set this to zero to find the maximum.

[latex] \displaystyle{\frac{1}{2}A-3\pi r^{2}=0 } [/latex]

Solving for [latex]r[/latex] we get

[latex] \displaystyle{ r=\sqrt{\frac{A}{6\pi}}=4.60659} [/latex]

Substitute this value for [latex]r[/latex] into equation c. to find the corresponding value for [latex]h[/latex]

[latex] \displaystyle{h=\frac{A}{2\pi r}-r=\frac{A}{2\pi}\sqrt{\frac{6\pi}{A}}-\sqrt{\frac{A}{6\pi}}=\sqrt{\frac{2A}{3\pi}}=9.21318 } [/latex]

I left out some simplification lines for clarity. Interesting that [latex]h[/latex] must be exactly [latex]2r[/latex], i.e. the height must be the same as the diameter for maximum volume.

Intergration calculating centroid of an area

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