Respuesta :
Answer:
The volume of the cone is shrinking at the rate [tex]100\pi[/tex] cubic inches / min.
Step-by-step explanation:
Formula:
- [tex]\frac{d}{dx}(uv)=v\frac{du}{dt}+u\frac{dv}{dt}[/tex]
- [tex]\frac{d}{dx}(u^n)=nu^{n-1}\frac{du}{dx}[/tex]
Given that,
The base of a cone is expanding at the rate of 1 in / min while it height is shrinking at the rate 3 in/ min.
i.e
[tex]\frac{dr}{dt}=1 \ in/min[/tex]
and
[tex]\frac{dh}{dt}= -3 \ in/min[/tex] [ since height is shirking]
The volume of a cone is
[tex]V=\frac13 \pi r^2h[/tex]
Differentiating with respect to t
[tex]\frac{dV}{dt} =\frac13\pi \{h \frac{d}{dt}(r^2)+r^2\frac{d}{dt}(h)\}[/tex]
[tex]\Rightarrow\frac{dV}{dt}=\frac13\pi (h.2r \frac{dr}{dt}+r^2\frac{dh}{dt})[/tex]
Now putting [tex]\frac{dr}{dt}=1 \ in/min[/tex] and [tex]\frac{dh}{dt}= -3 \ in/min[/tex]
[tex]\frac{dV}{dt}=\frac13\pi \{h.2r (1)+r^2(-3)\}[/tex]
[tex]\Rightarrow \frac{dV}{dt}=\frac13\pi (2hr -3r^2)[/tex]
To find the rate of changing volume at the moment when base radius is 30 in and the height is 40 in, we need to put r= 30 in and h=40 in.
[tex]\frac{dV}{dt}|_{r=30,h=40}=\frac13\pi \{2.40.30 -3(30)^2\}[/tex]
[tex]=-100\pi[/tex] cubic inches / min
The volume of the cone is shrinking at the rate [tex]100\pi[/tex] cubic inches / min.
