Respuesta :
Answer:
Step-by-step explanation:
We have volume of cone as
[tex]V=\frac{1}{3} \pi r^2 h[/tex]
and for a cone always r/h = constant
Given that r' = rate of change of radius = -7 inches/sec
(Negative sign because decresing)
V' =- 948 in^3/sec
Radius = 99 inches and volume = 525 inches
Height at this instant = [tex]\frac{525}{\frac{1}{3} \pi (99)^2} \\=\frac{0.1607}{\pi}[/tex]
Let us differentiate the volume equation with respect to t using product rule
[tex]V=\frac{1}{3} \pi r^2 h\\V' = \frac{1}{3} \pi[2rhr'+r^2 h']\\-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\[/tex]
[tex]-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\-948 = 33(3.14)(-2.25/3.14 + 99 h')\\-9.149=-0.72+99h'\\-8.429 = 99h'\\h' = 0.08514[/tex]
Rate of change of height = 0.08514 in/sec
The Rate at which the height of cone changing is [tex]0.085\;\text{inch/sec}[/tex]
Step-by-step explanation:
Given information:
The radius is decreasing at a constant rate of [tex]7\;\text{inch}/\text{sec.}[/tex]
The volume is decreasing at a rate of [tex]948\;\text{inch}^3/\text{sec.}[/tex]
The volume of cone is
[tex]V=\frac{1}{3}\pi r^2h[/tex]
At a instant the volume is [tex]525\; \text{inch}^3[/tex] and the radius is 99 inch.
At that instant :
The height of the cone is
[tex]=\frac{525}{\frac{1}{3} \pi (99^2) }[/tex]
[tex]=0.1607/\pi[/tex]
Now, differentiate the volume equation with respect to "[tex]\text t[/tex]"
[tex]V'=1/3\times \pi [2rhr'+r^2h']\\-948=1/3\times \pi [2(99)(-7)(0.1607/\pi)+99^2 \times h']\\-948=33(3.14)(-2.25/3.14)+99h'\\-8.429=99h'\\h'=0.08514[/tex]
Hence , The rate at which the height of cone changing is [tex]0.085\;\text{inch/sec}[/tex]
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