Question

The length of a solid cylindrical cord of elastic material is 32 inches. A circular cross section of the cord has radius 1/2 inch.The cord is stretched lengthwise at a constant rate of 18 inches per minute. Assuming that the cord maintains a cylindrical shape and a constant volume, at what rate is the radius of the cord changing one minute after the stretching begins? Indicate units of measure.

Answer 1

Answer:

-0..072 Inches Per Minutes

Step-by-step explanation:

Volume of a Cylinder, V = $\pi r^2 l$

Given l=32 Inches, r=0.5 Inches

V = $\pi*0.5^2*32=8\pi \:cubic \:inches$

$\frac{dV}{dt}=\pi r^2\frac{dl}{dt} +\pi l(2r)\frac{dr}{dt}$

Since the Volume is constant, $\frac{dV}{dt}=0$

After one minute,

• Length of the Cord=32+18=50 inches

Since the volume is constant

$V=\pi*r^2*50=8\pi\\r^2=\frac{8}{50} =\frac{4}{25} \\r=\sqrt{\frac{4}{25}} =\frac{2}{5}\\\text{Radius after 1 Minute}=\frac{2}{5} Inches$

Therefore:

$\frac{dV}{dt}=\pi r^2\frac{dl}{dt} +\pi l(2r)\frac{dr}{dt}\\0=\pi*(\frac{2}{5})^2*18+\pi*50*2(\frac{2}{5})\frac{dr}{dt}\\0=2.88\pi+40\pi \frac{dr}{dt}\\40\pi \frac{dr}{dt}=-2.88\pi\\40 \frac{dr}{dt}=-2.88\\\frac{dr}{dt}=-2.88 \div 40\\\frac{dr}{dt}=-0.072 Inches per minute$

Therefore, the radius of the cord is reducing at a rate of 0.072 Inches per minute.