Results 1 to 5 of 5

Math Help - Linear Approximation problems.....

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    24

    Linear Approximation problems.....

    1. Let f(t) be the weight (in grams) of a solid sitting in a beaker of water. Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time can be determined from the weight using the formula f'(t)=-3f(t)(5+(f(t)). If there are 4 grams of solid at t=2, estimate the amount of solid one second later.

    2. Use Linear Approximation to approximate 1.6^4 as follows: Let f(x)=x^4. The equation of the tangent line to f(x) at x=2 can be written in the form y=mx+b, where m=______ and b=______. Using this, the approximation of 1.6^4 is _______.

    3. The L.A. at x=0 of Sqrt(3+5x) can be written as A+Bx where A=______ and B=______.

    4. Use L.A. to approximate 1/1.003 as follows: Let f(x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 1.003. Then use this to estimate 1/1.003.

    5. The linearization at x=0 to sin(6x) is A+Bx . Compute A and B.


    For number 4 I've gotten that it's approximately 0.997017919, but it keeps saying that it's wrong. I have no clue how to figure out any of the others.
    Last edited by john11235; October 26th 2008 at 07:12 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    5) Have you considered actually answering the question? It asks for A and B and you have supplied only one value. How can that be right?

    Think about the most basic algebra and geometry. Remember the Point Slope form of a line? (y-y0) = m(x-x0). That's all you need to know.

    We have x0 = 0, that gives y0 = sin(6(0)) = sin(0) = 0

    Further, if f(x) = sin(6x), we have f'(x) = 6cos(6x) and immediately m = f'(0) = 6cos(6(0)) = 6cos(0) = 6(1) = 6.

    Putting it all together (y-y0) = m(x-x0) leads to (y-0) = 6(x-0). Simplifying a bit, we get y = 6x and we see that A = 0 and B = 6.

    You show us the next one. 4 - maybe?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    24
    Quote Originally Posted by TKHunny View Post
    5) Have you considered actually answering the question? It asks for A and B and you have supplied only one value. How can that be right?

    Think about the most basic algebra and geometry. Remember the Point Slope form of a line? (y-y0) = m(x-x0). That's all you need to know.

    We have x0 = 0, that gives y0 = sin(6(0)) = sin(0) = 0

    Further, if f(x) = sin(6x), we have f'(x) = 6cos(6x) and immediately m = f'(0) = 6cos(6(0)) = 6cos(0) = 6(1) = 6.

    Putting it all together (y-y0) = m(x-x0) leads to (y-0) = 6(x-0). Simplifying a bit, we get y = 6x and we see that A = 0 and B = 6.

    You show us the next one. 4 - maybe?
    Sorry, I hit the wrong button there.... I meant to say that I had tried number 4.

    But now it's not making any sense.... I have (Y-Yo)=(-1/1.003^2)(x-1.003), but what's the Yo supposed to be?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    You missed the really big hint! "Nice Point"

    x0 is NOT 1.003. You should be using the "Nice Point" for x0 and y0.

    Try x0 = 1 and 1/x0 = 1/1 = 1 = y0 so that you can approximate 1/1.003.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    24
    Quote Originally Posted by TKHunny View Post
    You missed the really big hint! "Nice Point"

    x0 is NOT 1.003. You should be using the "Nice Point" for x0 and y0.

    Try x0 = 1 and 1/x0 = 1/1 = 1 = y0 so that you can approximate 1/1.003.
    Oh ok.

    I got it now. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Approximation Problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 17th 2010, 05:10 AM
  2. Differential approximation problems
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 2nd 2009, 06:14 PM
  3. Linear approximation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 23rd 2008, 09:31 PM
  4. Replies: 3
    Last Post: January 2nd 2008, 03:47 PM
  5. wow linear approximation
    Posted in the Calculus Forum
    Replies: 11
    Last Post: November 4th 2007, 07:51 PM

Search Tags


/mathhelpforum @mathhelpforum