Respuesta :
Complete question is;
A 10 cm thick grindstone is initially 200 cm in diameter, and it is wearing away at a rate of 50 cm/hr. At what rate is its diameter decreasing?
Answer:
Diameter is decreasing at the rate of 5/(2πr) cm/hr
Step-by-step explanation:
We are told the stone is wearing away at a rate of 50 cm/hr. This means the volume is decreasing. Thus;
dV/dt = -50 cm/hr
Now, a grindstone is in the shape of a cylinder. Thus, volume of grindstone is;
V = πr²h
dV/dr = 2πrh
Now,to find the rate at which the diameter is decreasing, we'll write;
dr/dt = (dV/dt)/(dV/dr)
dr/dt = -50/(2πrh)
We are given;
Diameter = 200 cm
Radius; r = 200/2 = 100 cm
Thickness; h = 10 cm
Thus;
dr/dt = -50/(2π × r × 10)
dr/dt = -5/(2πr) cm/hr
The rate at which grindstone diameter decreases is [tex]-5/2\pi r \;{\rm cm/hr}[/tex] and this can be determined by using the given data.
Given :
A 10 cm thick grindstone is initially 200 cm in diameter and it is wearing away at a rate of 50 [tex]\rm cm^3/hr[/tex].
The following steps can be used in order to determine the rate at which grindstone diameter decreases:
Step 1 - According to the given data, the rate at which grindstone volume decreases is:
[tex]\dfrac{dV}{dt} = 50\;{\rm cm^3/hr}[/tex] --- (1)
Step 2 - The formula of the volume of the cylinder (grindstone) is given below:
[tex]V = \pi r^2 h[/tex]
Step 3 - Differentiate the above expression with respect to 'r'.
[tex]\dfrac{dV}{dr} = 2\pi r h[/tex] --- (2)
Step 4 - So, using the expression (1) and (2) the rate at which grindstone diameter decreases is:
[tex]\dfrac{dr}{dt}=\dfrac{\frac{dV}{dt}}{\frac{dV}{dr}}[/tex]
[tex]\dfrac{dr}{dt}=\dfrac{-50}{2\pi r \times 10}\\[/tex]
[tex]\dfrac{dr}{dt} = -\dfrac{5}{2\pi r }\; {\rm cm/hr}[/tex]
So, the rate at which grindstone diameter decreases is [tex]-5/2\pi r \;{\rm cm/hr}[/tex].
For more information, refer to the link given below:
https://brainly.com/question/12748872
