Posted by Mishaka on .
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 
Steve,
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 
Ford B.,
First, let's sketch what we can derive geometrically.
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(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:$$(12)^2=4y^2+25\\4y^2=14425=119\\y^2=\frac{119}{4}\\y=\frac{\sqrt{119}}2$$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 
Charylopos,
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 
Charylopos,
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 
Charylopos,
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 
Charylopos,
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(14425) = 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