Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the above. At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone?

We have [tex]\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}[/tex] and we want to find [tex]\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?[/tex] when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

[tex]\displaystyle V =\frac{1}{3} \pi (5h)^2 h[/tex]
[tex]\displaystyle V =\frac{1}{3} \pi \ 25h^3[/tex]

Differentiate this equation with respect to time t.

[tex]\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}[/tex]
[tex]\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}[/tex]

Plug known values into the equation and solve for dh/dt.

[tex]\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}[/tex]
[tex]\displaystyle 10 = 100 \pi \ \frac{dh}{dt}[/tex]

Divide both sides by 100π to solve for dh/dt.

[tex]\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}[/tex]
[tex]\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}[/tex]

The height of the cone is increasing at a rate of 1/10π cm per second.

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the above. At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone?​

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Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the above. At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone?