Calculus

posted by .

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