Tuesday

May 3, 2016
Posted by **Mishaka** on Sunday, January 15, 2012 at 6:02pm.

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**, Monday, January 16, 2012 at 12:00amyou 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.**, Thursday, August 13, 2015 at 3:39pmFirst, 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 -
**Charylopos**, Thursday, August 13, 2015 at 3:40pmeach 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**, Thursday, August 13, 2015 at 3:40pmA.

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**, Thursday, August 13, 2015 at 3:41pmA.

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**, Thursday, August 13, 2015 at 3:41pmPythagorous 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