The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At what rate is the volume of the cone changing when the radius is 134 in. and the height is 136 in.

Answer:

Volume of the cone is increasing at the rate [tex]9916\pi \frac{in^3}{s}[/tex].

Step-by-step explanation:

Given: The radius of a right circular cone is increasing at a rate of [tex]1.9[/tex] in/s while its height is decreasing at a rate of [tex]2.2[/tex] in/s.

To find: The rate at which volume of the cone changing when the radius is [tex]134[/tex] in. and the height is [tex]136[/tex] in.

Solution:

We have,

[tex]\frac{dr}{dt} =1.9 \:\text{in/s}[/tex], [tex]\frac{dh}{dt}=-2.2\:\text{in/s}[/tex], [tex]r=134 \:\text{in}[/tex], [tex]h=136\:\text{in}[/tex]

Now, let [tex]V[/tex] be the volume of the cone.

So, [tex]V=\frac{1}{3}\pi r^{2}h[/tex]

Differentiate with respect to [tex]t[/tex].

[tex]\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ][/tex]

Now, on substituting the values, we get

[tex]\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left ( 136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ][/tex]

[tex]\frac{dv}{dt} =\frac{1}{3}\pi\left [ -39503.2+69251.2 \right ][/tex]  

[tex]\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ][/tex]

[tex]\frac{dv}{dt} =9916\pi \frac{in^3}{s}[/tex]

Hence, the volume of the cone is increasing at the rate [tex]9916\pi \frac{in^3}{s}[/tex].

[tex] \frac{2}{6 } + \frac{1}{4} = [/tex]
how do we solve?
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