Jump to content

Assignment Help


randy17
 Share

Recommended Posts

First, you should have posted in the homework forum.

 

Second, few people on this forum will help you unless you show that you have at least tried to solve your homework. They want to know exactly what has you stumped, and will help get you past your misunderstanding.

 

Third, each different homework problem should be in a different thread in the homework forum. Otherwise, people get confused about which problem is being worked as different people help with the different problems.

 

I might be able to help you with some of these, but cannot help with all, as I haven't used calculus since the 1970s. But, there are people here who can, PhD professors I believe. I assure you, few of them will help until you show you are willing to try. Neither will I.

Edited by EdEarl
Link to comment
Share on other sites

Hey, do you need help for all 11 problems.

if so there is another basic problem.

Any way one question at a time choose the hardest.Read your text book, refer.......

Edited by daniton
Link to comment
Share on other sites

Hi there,

 

To get started with the first few, you'll just need to recall a few basic identities - knowing ax =eln a x, the product rule and the quotient rule should get you through those three.

 

Once you've done that, think about the rules you've used and the question with s(t) should look easy.

 

The next one is a little messy so I'd suggest that first you try to work out da/dx for a=a2+ln(x).

 

For the graphing problem, think about range, value at x=0 and limits for both ends of x.

 

For the last bit on that page - just work through it real slow, so long as you remember the basic properties of logarithms then you should be fine.

 

Good luck tho!

Link to comment
Share on other sites

  • 2 weeks later...

Problems a,b and c are pretty straight forward. [math]2^x/e^x[/math] is the same as 2^x times e to the power of negative x. With this, not only do you have the product rule, but the chain rule in each product as well. Product rule is a'b+b'a, you mark one multiplier as "a", and the other as "b". and do (derivative of "a" times normal "b"), then add (derivative of "b" times normal "a"), which would be [math]2^x/(ln2 *e^x) + -e^(-x) * 2^x[/math]. It's a similar circumstance for b and c, but if you'll recall, the derivative of lnx is 1/x, so you'd do 1/polynomial times the derivative of that polynomial. Uses the quotient rule, but you still do a similar thing and us chain rule, you do [math]((a' * b) - (b' * a))/(b^2)[/math], where "b" is the denominator and a is the numerator.

To find the slope at a certain point, just find the derivative of the function, then plug in the x value that you are suppose to find the slope at. I think you can complete the first page. The second page is related rates and optimization. For the first related rates problem a good thing to do is to mark down the "permanent", "instantaneous" and "what you want". What you want is [math]d(water level height)/(dt)[/math], or the rate that the water level is changing. What you need to do is come up with an equation that uses the variables given in the problem, simplify it, and differentiate it so that you get dw/dt (w=water level height). The "permanent" is the equation that describes the shape and how the volume of water fills it, which is what appears to be an equilateral triangular prism, so you need an equation for that prism involving the height of the water that fills it.

For the second one on the second page, just make an equation to model the profits and find the maximum by taking the derivative of the profit equation and seeing where the y value equals 0.

The third problem is related rates, you can solve it easily by labeling what I suggested before with marking what you want and the permanent and instantaneous and use the relationship of similar triangles.

Try the fourth one on your own if we can do these other ones.

Edited by SamBridge
Link to comment
Share on other sites

Sambridge and Randy17

 

for the first problem I would approach it thus

 

[latex]y=\frac{2^x}{e^x} = \left( \frac{2}{e} \right)^x[/latex]

 

[latex]\frac{dy}{dx}= \left( \frac{2}{e} \right) ^x log \left( \frac{2}{e} \right)[/latex]

 

[latex]\frac{dy}{dx}= \left( \frac{2}{e} \right)^x (log(2)-1)[/latex]

 

There is no need for the chain rule as 2/e is a constant. I haven't bothered to rearrange Sam's answer but from first glance I am not sure it is correct.

Link to comment
Share on other sites

Sambridge and Randy17

 

for the first problem I would approach it thus

 

[latex]y=\frac{2^x}{e^x} = \left( \frac{2}{e} \right)^x[/latex]

 

[latex]\frac{dy}{dx}= \left( \frac{2}{e} \right) ^x log \left( \frac{2}{e} \right)[/latex]

 

[latex]\frac{dy}{dx}= \left( \frac{2}{e} \right)^x (log(2)-1)[/latex]

 

There is no need for the chain rule as 2/e is a constant. I haven't bothered to rearrange Sam's answer but from first glance I am not sure it is correct.

I suppose you could treat it like that, but quotient rule or product rule should be applicable, I get an answer after all. I think one mistake I made though was putting the ln(2) in the denominator rather than the numerator. However, I still can obtain (2^*nl(2)*e^x - e^x2^x)/e^2x using the quotient rule, which actually reduces to what you said, since it equals 2^x(ln2-1)/e^x.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.