Solving for Zero Velocity in Calculus: 81t^2-16t^-2 = 0

[thomasrules]

i can't factor 0=81t^2-16t^-2

The question was to find when the velocity will be zero from s=27t^3+(16/t)+10

 

[Nylex]

Factorise by t^-2 maybe: 0 = (81t^4 - 16)t^-2. I'm not too sure though, HTH anyway.

 

[thomasrules]

i don't think i can find t that way

 

[The Bob]

i can't factor 0=81t^2-16t^-2

The question was to find when the velocity will be zero from s=27t^3+(16/t)+10

$$81t^2 - 16t^{-2} = 0$$ is what you say you have and you need to do what with it?

The Bob (2004 ©)

 

[cepheid]

Yeah okay, so you differentiated that equation for s and got the velocity as a function of time (looks like you differentiated correctly too!) So, when will the velocity equal zero?

You already set up the equation:

81t² - 16t^-2 = 0

81t² = 16t^-2

Why not divide both sides by t^-2?

81t⁴ = 16

t⁴ = 16/81

Notice that 2⁴ = 16 and 3⁴ = 81

So your equation becomes: t⁴ = (2/3)⁴
t = 2/3

You were right! It was easy calculus, followed by easy algebra!

 

[thomasrules]

This is the question:

A particle moving along a straight line will be s cm from a fixed point at time t seconds, where t>0 and s=27t^3+(16/t)+10
a) FInd when the velocity will be zero.

So what I did was take derivative and set s to 0 but then I'm stuck

 

[dextercioby]

Well,your derivative is the one you "painted" first.

How about
$$t^{2}=u$$

And instead of solving for "t",u solve for "u"...

Daniel.

 

[dextercioby]

Yeah okay, so you differentiated that equation for s and got the velocity as a function of time (looks like you differentiated correctly too!) So, when will the velocity equal zero?

You already set up the equation:

81t² - 16t^-2 = 0

81t² = 16t^-2

Why not divide both sides by t^-2?

81t⁴ = 16

t⁴ = 16/81

Notice that 2⁴ = 16 and 3⁴ = 81

So your equation becomes: t⁴ = (2/3)⁴
t = 2/3

You were right! It was easy calculus, followed by easy algebra!

Basically,what u did is wrong...The equation has 4 distinct solutions...The imaginaty ones can be thrown away,but the negative one,i don't see why??

Daniel.

 

[thomasrules]

dexter his answer was correct though...I understand his way...but not how you do it...

0=81u-16u^-1

 

[thomasrules]

Dudes but you got to help me with one horrible question:

It's the chapter Optimizing in Economics and Science and is so damn hard...

Your neighbours operate a successful bake shop. One of their specialties is a very rich whipped cream cake. They buy the cakes from a supplier who charges 6$ per cake, and they sell 200 cakes weekly at a price of 10$ each. Research shows that profit from the cake sales can be increased by increasing the price. Unfortunately, for every increase of 0.5$, sales will drop by seven.

a) What is the optimal sales price for the cake to obtain a maximal weekly profit?

Keep in mind guys that I'm not just giving you my HM to do. I have tried these question many times, and I suck.

 

[cepheid]

Basically,what u did is wrong...The equation has 4 distinct solutions...The imaginaty ones can be thrown away,but the negative one,i don't see why??

Daniel.

Because I made the assumption that we wanted to look for times after "t -naught", an assumption that was validated retroactively when thomas posted the actual question by the statement that t must be > 0!

What I did was at worst, incomplete, but not "wrong"! And it was *exactly* correct for the problem with the given restrictions. So stop being so goddamn pretentious in every single freakin thread!

 

[bross7]

It sounds stupid the way I am saying it, but it doesn't help to get down on yourself and go on about how you suck. Eventually you will get this stuff and asking is never wrong.

Start by breaking down the information that you are given. Sometimes it is easier to dig into a problem when you take the data out of the sentences.

Initially
cost to buy = $6
cost to sell = $10
#sold = 200

So initially each cake is sold for a profit of $4 (10-6).

They want to maximize the revenue they take in each week so set a variable to equal the change in price/loss of sales.

Let x be the # of $0.5 increases

Then you have to set up an equation with respect to your new condition.

f(x) = (price of each cake)(#sold per week)

 

[malco97]

Since 0 = 81t^2 - 16t^-2

Let y = t^-2

Therefore 0 = 81y^-1 - 16y
= 81 - 16y^2
= (9 - 4y)(9+4y)
So, y = 9/4 and y = -9/4

But y = t^-2, so 9/4 = t^-2
9t^2 = 4
t = +/- 2/3, -2/3 can be ignored assuming we are only interested in times after t=0.

Solutions regarding y = - 9/4 are complex so they can be ignored.

When the velocity is 0, the time is 2/3 s.

 

[thomasrules]

so I already had that...this is what i had

f(x)=(800+0.5x)(200-7x)

f(x)=160000-5500x-3.5x^2

 

[thomasrules]

i think that's wrong man

 

[bross7]

I'm sorry, I wasn't reading the question correctly. Completely ignore what I had initially said.

This problem takes the form:

Revenue = Profit - Cost
R(x) = (price sold at)(#sold) - (6)(#sold)

So your equation will end up being

$$R(x) = (10+0.5x)(200-7x) - (6)(200-7x)$$
$$R(x) = -3.5x^2 + 72x + 800$$

Then solve for R'(x) and set that to equal 0.
Solve for x and that should be your final answer.

 

[thomasrules]

BRoss lol that is sort of what i did except i put (4+0.5x) at the beginning instead of subtracting it at end.

But I still don't get the answer, try it out man.

0=-3.5x^2+72x+800

7x=72
x=10.******

WHICH IS WRONG...answer is 15$

 

[learningphysics]

BRoss lol that is sort of what i did except i put (4+0.5x) at the beginning instead of subtracting it at end.

But I still don't get the answer, try it out man.

0=-3.5x^2+72x+800

7x=72
x=10.******

WHICH IS WRONG...answer is 15$

x is not the price... it's the number of increases of $0.5.

The price is 10+0.5x = 10 + 0.5(10) = 15

 

[thomasrules]

Oooooooooooooooooooooooooppppppppppppppppppsssssssssssssssssssssssssssss