Results 1 to 6 of 6

Math Help - Please check my answers.

  1. #1
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823

    Please check my answers.

    Hi, I've gotta turn this in in an hour. I just want to know if I've worked the problems correctly.

    Problem (1):

    Initially there were 100 mg of a radiactive substance present. After 6 hours the mass decreased by 3%. If the rate of decay is proportional to the amount of substance present at any time, find the amount remaining after 12 hours.

    My answer: 100e^{\frac{\ln(.97)}{6}(24)}\approx84.9. I thought that k would be negative here. why isn't it?


    (2) :

    When interest is compounded continuously, the amount of money S increases at a rate proportional to the amount present at any time. That is \frac{dP}{dt}=rS.

    (i) Determine the amount accrued at the end of 5 years when 5000 is deposited in a savings account paying 5.75% annual interest compounded continuously.
    Answer: about $6,665.45

    (ii) How long will it take for the amount to double?
    Answer: A little over 12 years.

    (iii) Compare the amount obtained in(i) with the bakance in the accouont if the interest were compounded quarterly.
    Answer: $6,651.82. The difference amounts to a family meal taken at mcdonald's

    (3) Solve: (x^2-y^2)dx+xydy=0
    Answer: y^2=\frac{1}{2}(cx^{-2}-x^2)

    (4) Solve (x-y)dx-xdy=0
    Answer: x(x-2y)=c

    (5) Solve \frac{dy}{dx}-5y=\frac{-5}{2}xy^3
    Answer: y^{-2}=\frac{1}{2}x-\frac{1}{20}+e^{-10x}c

    And I need help with this one...

    Let P(t) be the balance in your retirement account t years after retirement from which W is the amount withdrawn every year that pays interest at the rate of r percent per year, compounded continuously. After some consideration you will need $50,000 each year to live on after you retire and that you plan on living 30 years after your retirement. Assuming that your retirement account will earn 5% interest while you are taking out %50,000, how much money must be in the retirement account after you retire?

    This problem is just worded funny. Can some help me understand what it is saying and help me set it up?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by VonNemo19 View Post
    Let P(t) be the balance in your retirement account t years after retirement from which W is the amount withdrawn every year that pays interest at the rate of r percent per year, compounded continuously. After some consideration you will need $50,000 each year to live on after you retire and that you plan on living 30 years after your retirement. Assuming that your retirement account will earn 5% interest while you are taking out %50,000, how much money must be in the retirement account after you retire?

    This problem is just worded funny. Can some help me understand what it is saying and help me set it up?
    \text{Payment}=50,000 \ \ \ \text{Interest}=i=.05

    \displaystyle\text{Present Value of Annuity}=\frac{\text{PMT}}{i}\cdot\left(1-\frac{1}{(1+i)^{365t}}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by VonNemo19 View Post

    Problem (1):

    Initially there were 100 mg of a radiactive substance present. After 6 hours the mass decreased by 3%. If the rate of decay is proportional to the amount of substance present at any time, find the amount remaining after 12 hours.

    My answer: 100e^{\frac{\ln(.97)}{6}(24)}\approx84.9. I thought that k would be negative here. why isn't it?
    \displaystyle \frac{dA}{dt}=-kA\Rightarrow \int\frac{dA}{A}=-\int kdt\Rightarrow \ln|A(t)|=-kt+C\Rightarrow A(t)=C_1e^{-kt}

    A(0)=C_1=100

    A(t)=100e^{-kt}

    \displaystyle A(6)=100e^{-6k}=97\Rightarrow k=-\frac{\ln\left(\frac{97}{100}\right)}{6}

    \displaystyle A(t)=100e^{\frac{t\ln\left(\frac{97}{100}\right)}{  6}}

    Remember take the limit

    \displaystyle\lim_{t\to\infty}100e^{\frac{t\ln\lef  t(\frac{97}{100}\right)}{6}}=\cdots

    If your k is correct, the limit should approach 0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by VonNemo19 View Post
    (2) :

    When interest is compounded continuously, the amount of money S increases at a rate proportional to the amount present at any time. That is \frac{dP}{dt}=rS. Either the S should be a P or the P should be a S

    (i) Determine the amount accrued at the end of 5 years when 5000 is deposited in a savings account paying 5.75% annual interest compounded continuously.
    Answer: about $6,665.45 Correct

    (ii) How long will it take for the amount to double?
    Answer: A little over 12 years. Correct

    (iii) Compare the amount obtained in(i) with the bakance in the accouont if the interest were compounded quarterly.
    Answer: $6,651.82. The difference amounts to a family meal taken at mcdonald's Correct
    Everything is fine except your letters in the DE
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by VonNemo19 View Post
    (3) Solve: (x^2-y^2)dx+xydy=0
    Answer: y^2=\frac{1}{2}(cx^{-2}-x^2)
    I obtained a different solution for 3.

    This equation can be made exact by \displaystyle x^{-3}

    x^{-3}(x^2-y^2)dx+x^{-2}ydy=0

    \displaystyle\frac{\partial M}{\partial y}=\frac{-2y}{x^3}=\frac{\partial N}{\partial x}=\frac{-2y}{x^3}

    \displaystyle\frac{\partial f}{\partial y}=x^{-2}y\Rightarrow f(x,y)=\frac{y^2}{2x^2}+h(x)\Rightarrow f_x=-x^{-3}y^2+h'(x)=x^{-1}-x^{-3}y^2\Rightarrow h'(x)=x^{-1}

    h(x)=\ln|x|

    \displaystyle C=\frac{y^2}{2x^2}+\ln|x|\Rightarrow y^2=2x^2(C-\ln|x|)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by VonNemo19 View Post
    (4) Solve (x-y)dx-xdy=0
    Answer: x(x-2y)=c
    This one is exact.

    \displaystyle\frac{\partial M}{\partial y}=-1=\frac{\partial N}{\partial x}=-1

    \displaystyle\frac{\partial f}{\partial y}=-x\Rightarrow f(x,y)=-xy+h(x)\Rightarrow f_x=-y+h'(x)=x-y\Rightarrow h'(x)=x\Rightarrow h(x)=\frac{x^2}{2}

    \displaystyle c=-xy+\frac{x^2}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I want check my answers
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 2nd 2010, 08:13 PM
  2. Can someone please check my answers?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 2nd 2010, 10:19 AM
  3. Getting better - but still need to check my answers!!
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: April 2nd 2008, 05:45 PM
  4. Please check my answers
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 26th 2006, 08:53 AM
  5. I just want to check my answers, can you help me out?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 27th 2005, 08:49 AM

Search Tags


/mathhelpforum @mathhelpforum