Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Tangent line approximation with sinx/x + siny/y = C

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose the point (pi/3, pi/4) is on the curve sinx/x + siny/y = C, where C is a constant. Use the tangent line approximation to find the y-coordinate of the point on the curve with x-coordinate pi/3 + pi/180.

    2. Relevant equations

    TLA: f(a) + f'(a)(x-a)

    Where a is the known value of x and x is the value you are estimating the output of.

    3. The attempt at a solution

    Before differentiating I figured that I could replace siny/y with siny * y^-1.
    I got here:

    ((x)(cosx) – sinx)/x2 + y’(siny)(-y-2) + cosy(y’)(y-1) = 0

    which I THINK can be simplified to:

    ((x)(cosx) – sinx)/x2 + y’(siny*-y-2 + cosy*y-1)

    dunno where to go with it from there D:

    MAYBE, here's what I attempted:

    ((x)(cosx) - sinx)/x2 = -y’(siny*-y-2 + cosy*y-1)
    -((((x)(cosx) - sinx)/x2)/(siny*-y-2 + cosy*y-1)) = y'

    But that doesn't seem right. The question gave the curve sinx/x + siny/y = C with the point (pi/3, pi/4) given. Substituting I got C = 1.7273097. However using tangent line approximation with the derivative I calculated I got the y coordinate there to be 418.879:

    f(a) + f’(a)(x – a)
    f(π/3) + f’(π/3)(π/180)
    1.7273097 + 24000(π/180)

    I know I could just do normal algebra for this instead of calculus but the instructions say to use tangent line approximation so I think I'd get points docked. I am pretty sure though the answer I'm after is y = 0.905801 based on just solving for the values using C = 1.7273097 and x = pi/3 + pi/180... pretty far off from 418.879.

    Someone please help, I've got no idea what to do here.

    If it helps get the help any quicker I don't really care about a pretty, formatted answer... I really need help on this soon so whatever helps.
     
    Last edited: Dec 18, 2011
  2. jcsd
  3. Dec 18, 2011 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hello franklingroye. Welcome to PF.

    I agree with your result for C.

    What is your result for y' when x = π/3 and y = π/4 ?

    I got y' ≈ -0.000876215 there.

    The way you are using the tangent line approximation is wrong.

    y = f(x), so f(π/3) = π/4
     
  4. Dec 18, 2011 #3
    Goodness, I can't believe I managed to mess up with f(x) haha.

    I'm not sure how to really get y' out of this is the issue I am having. If I leave it as-is instead of changing it to siny*y-1:

    ((x)(cosx) – sinx)/x2 + y'((y)(cosx) – siny)/y2
    ((x)(cosx) – sinx)/x2 = -y'((y)(cosy) – siny)/y2
    [((x)(cosx) – sinx)/x2]/(y)(cosy) – siny)/y2 = -y'
    -[((x)(cosx) – sinx)/x2]/(y)(cosy) – siny)/y2 = y'
    -((((pi/3)(cos(pi/3)) - sin(pi/3))/(pi/3)^2)/((((pi/4))(cos(pi/4)) - sin(pi/4))/(pi/4)^2)) = -1.269321

    I've got no idea where I'm screwing up

    I tested the TLA equation with y' = -0.000876215 and it's correct but now I just need to figure out how to get there. :\ I'm obviously messing up somewhere in the differentiation process.
     
  5. Dec 18, 2011 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I checked your work to here and it looked good.
    ((x)(cosx) - sinx)/x2 = -y’(siny*(-y-2) + cosy*y-1)​

    Solving that for y', I got:
    [itex]\displaystyle y'=-\frac{y^2\left(x\cos(x)-\sin(x)\right)}{x^2\left(y\cos(y)-\sin(y)\right)}[/itex]​

    And I see that I had a sign error when I calculated y' at the given point.

    It should have been y' ≈ 0.00728974

    (Yes, it's positive.)
     
  6. Dec 18, 2011 #5
    I don't get what's going on here, using my calculator to simply plug in pi/3 and pi/4 for x/y in your derivative I end up with 0.15257. ,-,
     
  7. Dec 18, 2011 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Check parentheses.

    After rechecking my algebra, I now get y' ≈ -1.26932145 ; back to a negative, and it should be negative when considering the values of x & y used in
    [itex]\displaystyle y'=-\frac{y^2\left(x\cos(x)-\sin(x)\right)}{x^2\left(y\cos(y)-\sin(y)\right)}[/itex]​
    For the values of x & y used here, cos(x)=1/2, sin(x)=(√3)/2, cos(y)=sin(y)=(√2)/2 .
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook