Related Rates Solution 16

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SOLUTION 16: Consider the given side view of the conical tank of height

$10$ meters and base radius

$8$ meters.

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Consider the cone of water of depth

$h$ meters and the right triangle with legs

$r$ and

$h$ in the side view of the conical tank. Assume that lengths

$r$ and

$h$ are both functions of time

$t$ .

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RECALL: The volume of a right circular cone of height

$h$ and base radius

$r$ is

$V = \displaystyle{ 1 \over 3 } \pi r^2 h$

GIVEN:

$\ \ \ \displaystyle{ dV\over dt } = \pi \ m^3/min.$

FIND:

$\ \ \ \displaystyle{ dh \over dt }$ when

$\ \$ a.)

$\ h=1 \ m.$

$\ \$ b.)

$\ h=9 \ m.$

First consider the two given similar right triangles. Use them to write

$r$ in terms of

$h$ .

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Using similar triangles we can conclude that

$\displaystyle{ r \over h } = { 8 \over 10 } \ \ \ \ \longrightarrow$

$\displaystyle{ r \over h } = { 4 \over 5 } \ \ \ \ \longrightarrow$

$r = \displaystyle{ 4 \over 5 }h$

The volume equation can now be rewritten as

$V = \displaystyle{ 1 \over 3 } \pi \Big\{ { 4 \over 5 }h \Big\}^2 h \ \ \ \ \longrightarrow$

$V = \displaystyle{ 16 \over 75 } \pi h^3$

Differentiating the volume equation, we get

$D \{ V \} = D\{ \displaystyle{ 16 \over 75 } \pi h^3 \} \ \ \ \ \longrightarrow$

$\displaystyle{ dV \over dt } = \displaystyle{ 16 \over 75 } \pi \cdot 3h^2 { dh \over dt } \ \ \ \ \longrightarrow$

$(**) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle{ dV \over dt } = \displaystyle{ 16 \over 25 } \pi h^2 { dh \over dt } \ \ \ \ \longrightarrow$

$\Big($ a.) Now let

$\displaystyle{ dV \over dt } = \pi$ and

$\ h=1 . \Big)$

$\pi = \displaystyle{ 16 \over 25 } \pi (1)^2 { dh \over dt } \ \ \ \ \longrightarrow$

$\Big($ Divide both sides of the equation by

$\pi. \Big)$

$1 = \displaystyle{ 16 \over 25 } { dh \over dt } \ \ \ \ \longrightarrow$

$\displaystyle{ dh \over dt } = { 25 \over 16 } =1.5625 \ meters/min.$

$\Big($ b.) Now let

$\displaystyle{ dV \over dt } = \pi$ and

$\ h=9$ in equation (**).

$\Big)$

$\pi = \displaystyle{ 16 \over 25 } \pi (9)^2 { dh \over dt } \ \ \ \ \longrightarrow$

$1 = \displaystyle{ 16 \over 25 } (81) { dh \over dt } \ \ \ \ \longrightarrow$

$1 = \displaystyle{ 1296 \over 25 } { dh \over dt } \ \ \ \ \longrightarrow$

$\displaystyle{ dh \over dt } = { 25 \over 1296 } \approx 0.019 \ meters/min.$

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