Question

How do you find the rate at which the volume of a cone changes with the radius is $40$ inches and the height is $40$ inches, where the radius of a right circular cone is increasing at a rate of $3$ inches per second and its height is decreasing at a rate of $2$ inches per second?

Answer 1

Hint: In this question, we have to find the rate at which the volume of a cone. By using the formula for difference and putting the given values. On doing some simplification we get the required answer

Formula used: Circular Cone Formulas in terms of radius r and height h: $V = \pi {r^2}\dfrac{h}{3}$ , where r denotes the radius of the base of the cone, and h denotes the height of the cone.
Volume of a cone: $V = \pi {r^2}\dfrac{h}{3}$

Complete step-by-step solution:
For a cone with radius $40$ inches which is increasing at the rate of $3$ inches per second, and with a height of $40$ inches which is decreasing at the rate of $2$ inches per second, the rate at which the volume changes will be found like,
We know that $V = \pi {r^2}\dfrac{h}{3}$
Differentiate implicitly with respect to $t$
$ \Rightarrow \dfrac{d}{{dt}}(V) = \dfrac{d}{{dt}}\left( {\dfrac{1}{3}\pi {r^2}h} \right)$
Simplifying it further,
$ \Rightarrow = \dfrac{1}{3}\pi \dfrac{d}{{dt}}({r^2}h)$
We'll need the product rule on the right.
$ \Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{1}{3}\pi \left[ {2r\dfrac{{dr}}{{dt}}h + {r^2}\dfrac{{dh}}{{dt}}} \right]$
(I use the product rule in the form$((FS)' = F'S + FS')$)
Substitute the given information and do the arithmetic.
$ \Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{1}{3}\pi \left[ {2(40)(3)(40) + {{(40)}^2}( - 2)} \right]$
Multiplying the values in the brackets,
$ \Rightarrow = \dfrac{1}{3}\pi \left[ {6{{(40)}^2} - 2{{(40)}^2}} \right]$
Simplifying it further,
$ \Rightarrow = \dfrac{1}{3}\pi \left[ {4{{(40)}^2}} \right]$
Therefore, the final value will be,
$ \Rightarrow \dfrac{{6400\pi }}{3}$cu in/s

Therefore, the rate at which the volume will change for a cone with the radius is $40$ inches and the height is $40$ inches is and where the radius of a right circular cone is increasing at a rate of $3$ inches per second and its height is decreasing at a rate of $2$ inches per second is $\dfrac{{6400\pi }}{3}$ cu in/s

Note: Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre. A cone is a solid with a circular base. It has a curved surface which tapers (i.e., decreases in size) to a vertex at the top. The height of the cone is the perpendicular distance from the base to the vertex. The volume of a right cone is equal to one-third the product of the area of the base and the height.