Post a New Question

Calculus (Normal Line, please check my work)

posted by on .

The slope of the line normal to the graph of 4 sin x + 9 cos y = 9 at the point (pi, 0) is:
Derivative: 4cosx - 9siny(dy/dx) = 0
(dy/dx) = (-4cosx) / (-9siny)
(dy/dx) = (4) / 0
Normal line = -1 / (4/0)
Does this mean that the slope of the normal line is undefined, or did I do my calculations incorrectly???

  • Calculus (Normal Line, please check my work) - ,

    Looks to me like the slope is vertical at (pi,0)
    Therefore the normal is horizontal
    dy/dx of normal = 0

  • Calculus (Normal Line, please check my work) - ,

    One branch of the graph is an oval, and at (pi,0) does indeed have a vertical tangent. So, the normal line has zero slope.

    Line is just y=0

    go by wolfram dot com and ask it to graph your equation. You'll see you were right.

  • Calculus (Normal Line, please check my work) - ,

    For the tangent line at (pi, 0), I find that the slope is 4/0, which is undefined. So, wouldn't this make the slope of the normal line undefined as well??? Did I miss something?

  • Calculus (Normal Line, please check my work) - ,

    even better, have wolfram plot

    z = 4*sin(x) + 9*cos(y)

    and check "show contour lines"

    One of those contour lines will be at z=9, and will look show why your graph looks the way it does.

  • Calculus (Normal Line, please check my work) - ,

    Yes, you are missing something. A vertical line has undefined (infinite) slope.
    However the normal to that vertical line is simply a line with zero slope, a horizontal line.

  • Calculus (Normal Line, please check my work) - ,

    Graph it to see as Steve suggested.
    By the way Steve, did you see the message from Bob Pursley?

  • Calculus (Normal Line, please check my work) - ,

    Using strictly the derivative (because wolfram isn't working for me), how can you prove that the slope of the normal line is 0?

  • Calculus (Normal Line, please check my work) - ,

    -1/oo

    is very very small :)

  • Calculus (Normal Line, please check my work) - ,

    Where did you get infinity? Using differentiation, I found that:
    dy/dx = (-4cosx) / (9(-siny))
    When I put (pi, 0) into this equation, the denominator is 0, making the slope of the tangent line undefined. And since the slope of the normal line is -1/(slope of the tangent line), the normal line's slope would also be undefined. Is this is wrong, could you please show me step-by-step (using derivatives, please), how you came to your answer?

  • Calculus (Normal Line, please check my work) - ,

    Undefined means very large magnitude, like infinite.

    1/1 = 1
    1/.1 = 10
    1/.01 = 100
    .
    .
    .
    1/10^-6 = 1,000,000

    1/10^-10 = 10,000,000,000
    etc
    as the denominator goes to zero, the term goes to infinity

  • Calculus (Normal Line, please check my work) - ,

    If dy/dx is very very large
    then -1/(dy/dx) is very very small

  • Calculus (Normal Line, please check my work) - ,

    I apologize if I seem like I don't understand what you are trying to explain, but you have really confused me on what I thought was a more simple problem. I would really appreciate it if you could check my original answer using differentiation, instead, as this is the method I used to begin with.

  • Calculus (Normal Line, please check my work) - ,

    I did.
    You correctly found that the slope of the function was undefined or infinitely large at the desired point.
    You failed to make the connection that "undefined" means "huge". 1/zero is undefined and is the limit of 1/a very small number which is huge.
    If the slope of the function is huge, the slope of the normal is tiny.
    if m = 1/zero
    then m' = -1/m = -zero/1 = 0

  • Calculus (Normal Line, please check my work) - ,

    I think I may be understanding what you are saying. So, to put this in very simple turns, rather than assuming that -1 / (4/0) was just undefined, I should have taken it out farther; for example, dividing by a fraction is the same as multiplying by its reciprocal, which would be this:
    -1 x (0/4) = -1 x 0 = 0
    This better simplifies the answer for me and explains why the slope of the normal line is zero. Thanks!

  • Calculus (Normal Line, please check my work) - ,

    Sure, that is a perfectly good way to look at it.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question