Intergration calculating centroid of an area

TheBody

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

Hedgecutter

Bad wifi.

TheBody

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

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

Hedgecutter

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

TheBody

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

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

Third time lucky

TheBody

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

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

Well done. Keep up the hard work.

Hedgecutter

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

Hedgecutter

Here's the question

Hedgecutter

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

TheBody

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

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

dlouth15

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

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

dlouth15

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

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

My solutions

Hedgecutter

next question

dlouth15

Hedgecutter wrote: »

next question

This one looks OK.

dlouth15

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.