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March 26, 2017

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A piece of elastic is attached to two nails on a flat board, with a button attached to the midpoint of the elastic. The nails are 5 cm apart. You stretch the elastic by pulling the button along the board in a direction that is perpendicular to the line between the nails.
A. Fnd an equation that relates the total length of the elastic x to the distance y that the button has moved.
B. You pull the button at a constant 3 cm/sec. Find the rate at which the length of the elastic is increasing when it is 12 cm long.

  • Calculus - ,

    you have two right triangles of height y

    2.5^2 + y^2 = (x/2)^2
    or
    25 + 4y^2 = x^2

    8y dy/dt = 2x dx/dt
    when x = 12, y = 5.454

    8(5.454) = 2(12) dx/dt
    dx/dt = 1.818

  • Calculus - ,

    First, let's sketch what we can derive geometrically.
    |dw:1355290130364:dw|
    (A) Given we know those two triangles are right, we can relate the hypotenuse \(\frac12x\) to the altitude \(y\) and base \(\frac52\) using the Pythagorean theorem, which we can rearrange to yield an adequate relation:$$\left(\frac12x\right)^2=y^2+\left(\frac52\right)^2\\\frac14x^2=y^2+\frac{25}4\\x^2=4y^2+25$$
    (B) We're given that the button is moving at a rate of 3 cm/s, which can be expressed using a time derivative as \(\frac{dy}{dt}=3\). We're told that the elastic (at the instant we're interested in) is 12 cm long, i.e. \(x=12\); given this, we can determine the distance of the button from its initial position with relative ease:$$(12)^2=4y^2+25\\4y^2=144-25=119\\y^2=\frac{119}{4}\\y=\frac{\sqrt{119}}2$$Let's use implicit differentiation on our formula above to relate the rates of elongation:$$2x\frac{dx}{dt}=8y\frac{dy}{dt}\\24\frac{dx}{dt}=12\sqrt{119}\\2\frac{dx}{dt}=\sqrt{119}\\\frac{dx}{dt}=\frac{\sqrt{119}}2\approx5.4544$$

  • Calculus - ,

    each side of the elastic is the hypotenuse of a triangle with legs 5/2 and y, so

    x = 2√(2.5^2 + y^2)

    dx/dt = 2y/√(2.5^2+y^2) dy/dt
    when x=12, y=5.45, so
    dx/dt = 2(5.45)/6 (-3) = -1/5.45 = -0.18 cm/s

  • Calculus - ,

    A.
    x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem)

    B.
    x^2 = 4y^2 + 25
    => 2x dx/dt = 8y dy/dt
    => dx/dt = 4 (y/x) dy/dt
    When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm
    => dx/dt = 4 (2.727/12) x 3 cm/s
    = 2.727 cm/s.

  • Calculus - ,

    A.
    x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem)

    B.
    x^2 = 4y^2 + 25
    => 2x dx/dt = 8y dy/dt
    => dx/dt = 4 (y/x) dy/dt
    When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm
    => dx/dt = 4 (2.727/12) x 3 cm/s
    = 2.727 cm/s.

  • Calculus - ,

    Pythagorous Theorem,

    z^2 = y^2 + x^2, where y = 2.5 cm and x is the legth of the pull.

    substituting z^2 = 2.5^2 +x^2, z =sqrt(x^2 +6.25)

    dz/dx (2z) = 2x

    dz/dx = (x/z) ......... (a)

    but dz/dt = dz/dx * dx/dt

    Thus, dz/dt = (x/z) * dx/dt = x/z * 3

    when z = 12 , y = 5, x= sqrt(144-25) = sqrt(119)

    Thus dz/dt = sqrt(119)/12 * 3
    = sqrt(119)/4 =10.91 /4 =2.73 cm/sec

    Rate of increase in length of elastic = 2.73 cm/sec

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