Apc.2.8.1 ap vertical circular cylinder related rates

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SUMMARY

The discussion focuses on calculating the rate of increase of the lateral surface area of a vertical circular cylinder with a radius \( r \) and height \( h \) that both increase at a constant rate of 2 ft/sec. The formula derived for the lateral surface area \( A \) is \( A = 2\pi rh \). By applying the chain rule, the rate of change of the surface area with respect to time is determined to be \( \frac{dA}{dt} = 4\pi(r + h) \). This conclusion confirms that the lateral surface area increases at a rate dependent on the sum of the radius and height, both increasing at the same rate.

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  • Familiarity with the formula for the lateral surface area of a cylinder
  • Knowledge of differentiation techniques
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karush
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$\tiny{2.8.1}$

The vertical circular cylinder has radius r ft and height h ft.
If the height and radius both increase at the constant rate of 2 ft/sec,
Then what is the rate at which the lateral surface area increases?
\een
$\begin{array}{ll}
a&4\pi r\\
b&2\pi(r+h)\\
c&4\pi(r+h)\\
d&4\pi rh\\
e&4\pi h
\end{array}$
ok here is my setup
\begin{array}{lll}
\textit{given rates}
&\dfrac{dr}{dt}=2 \quad \dfrac{dh}{dt}=2
&(1)\\ \\
\textit{surface area eq}
&2\pi rh
&(2)\\ \\
\end{array}
so far
 
Last edited:
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$\dfrac{dA}{dt} = 2\pi\left(r \cdot \dfrac{dh}{dt} + h \cdot \dfrac{dr}{dt} \right)$
 
$\dfrac{dA}{dt} = 2\pi\left(r \cdot \dfrac{dh}{dt} + h \cdot \dfrac{dr}{dt} \right)
=2\pi(2r+2h)=4\pi(r+h)$
 

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