Results 1 to 4 of 4

Math Help - mathematical model

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    134

    mathematical model

    Can anyone help me with this ? Thank you in advance. I really need the answers because I will report this. Sorry to be of bother, I really have no idea because this hasn't been taught yet and I cant understand the book.

    1. A 20 ft. ladder leans against a vertical wall. If the bottom of the ladder is pulled away at the rate of 2 ft per sec, write a formula f(t), the distance of the top of the ladder from the floor as a function of t.

    2. Let P be the perimeter of an equilateral triangle. Write a formula A(P), the area of the triangle as function of the perimeter.

    3. One of the legs of a right triangle has a length of 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,654
    Thanks
    1478
    Quote Originally Posted by reiward View Post
    Can anyone help me with this ? Thank you in advance. I really need the answers because I will report this. Sorry to be of bother, I really have no idea because this hasn't been taught yet and I cant understand the book.

    1. A 20 ft. ladder leans against a vertical wall. If the bottom of the ladder is pulled away at the rate of 2 ft per sec, write a formula f(t), the distance of the top of the ladder from the floor as a function of t.

    2. Let P be the perimeter of an equilateral triangle. Write a formula A(P), the area of the triangle as function of the perimeter.

    3. One of the legs of a right triangle has a length of 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse.
    2. The perimeter of an equilateral triangle of length l will be 3l.

    So P = 3l or l = \frac{P}{3}.


    For any triangle, the area can be found using Heron's formula

    A = \sqrt{s(s - a)(s - b)(s - c)}, where s = \frac{a + b + c}{2}.

    Since it's an equilateral triangle, a = b = c = l.

    So s = \frac{3l}{2} and the area is

    A = \sqrt{\frac{3l}{2}\left(\frac{3l}{2} - l\right)\left(\frac{3l}{2} - l\right)\left(\frac{3l}{2} - l\right)}

     = \sqrt{\frac{3l}{2}\left(\frac{l}{2}\right)^3}

     = \sqrt{\frac{3l}{2}\left(\frac{l^3}{8}\right)}

     = \sqrt{\frac{3l^4}{16}}

     = \frac{\sqrt{3}l^2}{4}

     = \frac{\sqrt{3}\left(\frac{P}{3}\right)^2}{4}

     = \frac{\frac{\sqrt{3}P^2}{9}}{4}

     = \frac{\sqrt{3}P^2}{36}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,971
    Thanks
    1637
    Quote Originally Posted by reiward View Post
    Can anyone help me with this ? Thank you in advance. I really need the answers because I will report this. Sorry to be of bother, I really have no idea because this hasn't been taught yet and I cant understand the book.

    1. A 20 ft. ladder leans against a vertical wall. If the bottom of the ladder is pulled away at the rate of 2 ft per sec, write a formula f(t), the distance of the top of the ladder from the floor as a function of t.
    The floor, wall, and ladder form a right triangle with the length of the ladder as hypotenuse. Use the Pythagorean theorem to write an equation connecting those. If then the bottom leg of the triangle is given by a= a_0+ 2t where a_0 is the initial distance from the wall to the bottom of the ladder and t is the time in seconds.

    2. Let P be the perimeter of an equilateral triangle. Write a formula A(P), the area of the triangle as function of the perimeter.

    3. One of the legs of a right triangle has a length of 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse.
    Again, a right triangle. Dropping the altitude from the right angle to the hypotenuse divides that right trangle into two right triangles, both of which are similar to the original triangle. Pythagorean theorem and similar triangles.

    For all of these, the first thing you should have done was draw a picture. Then use what you know about triangles from geometry.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    693
    dHello, reiward!

    HallsofIvy gave you an excellent game plan for Problem #3.
    Did you follow it?


    3. One of the legs of a right triangle has a length of 4cm.
    Express the length of the altitude perpendicular to the hypotenuse
    as a function of the length of the hypotenuse.
    Code:
                  C
                  o
                 *| *
                * |   *
               *  |     *
            4 *   |y      *
             *    |         *
            *     |           *
           *      |             *
        A o * * * o * * * * * * * o B
          :   x   D      h-x      :
          : - - - - h - - - - - - :

    We have right triangle ABC,\;C = 90^o,\;AC = 4.

    Let h = AB, the hypotenuse.

    Let y = CD, atitude to hypotenuse.

    Let x = AD,\;\text{then: }h-x \:=\:DB


    We find that: . \Delta ADC \:\sim \: \Delta CDB \:\sim\:\Delta ACB

    Hence: . \dfrac{y}{x} \:=\:\dfrac{h-x}{y} \quad\Rightarrow\quad y^2 \:=\:hx-x^2 .[1]


    In \Delta ADC\!:\;x^2+y^2 \:=\:4^2 \quad\Rightarrow\quad x \:=\:\sqrt{16-x^2}


    Substitute into [1]: . y^2 \:=\:h\sqrt{16-y^2} - (16-y^2)

    . . y^2 \:=\:h\sqrt{16-y^2} - 16 + y^2 \quad\Rightarrow\quad h\sqrt{16-y^2} \:=\: 16

    . . \sqrt{16-y^2} \:=\:\dfrac{16}{h} \quad\Rightarrow\quad 16 - y^2 \:=\:\dfrac{256}{h^2}

    . . y^2 \:=\:16 - \dfrac{256}{h^2} \:=\:\dfrac{16h^2 - 256}{h^2} \:=\:\dfrac{16(h^2-16)}{h^2}


    Therefore: . y \;=\;\dfrac{4\sqrt{h^2-16}}{h}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. mathematical model
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: February 23rd 2009, 06:57 PM
  2. Mathematical Model
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: February 17th 2009, 10:25 AM
  3. mathematical model
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: November 12th 2007, 09:57 PM
  4. Challenging Mathematical Model
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: November 7th 2007, 08:02 PM
  5. Mathematical Model of a Catenary
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 19th 2007, 07:42 AM

Search Tags


/mathhelpforum @mathhelpforum