Mixing
Rate of change of cross-section of cylinder
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I got this task on my calculus class and I got stuck at process of figuring it out
A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.
What I know:
Volume is unknown and constant
Rate of change of diameter is $$dfrac {dD}{dt} = 0,2text{cm}/text{sec}$$
Rate of change of height is $$dfrac {dh}{dt} = -0,4text{cm}/text{sec}$$
Trying to find $dfrac{dD}{dt}$ by using formula $$V=pi r^2h$$
calculus
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
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add a comment |
$begingroup$
I got this task on my calculus class and I got stuck at process of figuring it out
A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.
What I know:
Volume is unknown and constant
Rate of change of diameter is $$dfrac {dD}{dt} = 0,2text{cm}/text{sec}$$
Rate of change of height is $$dfrac {dh}{dt} = -0,4text{cm}/text{sec}$$
Trying to find $dfrac{dD}{dt}$ by using formula $$V=pi r^2h$$
calculus
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
$endgroup$
add a comment |
$begingroup$
I got this task on my calculus class and I got stuck at process of figuring it out
A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.
What I know:
Volume is unknown and constant
Rate of change of diameter is $$dfrac {dD}{dt} = 0,2text{cm}/text{sec}$$
Rate of change of height is $$dfrac {dh}{dt} = -0,4text{cm}/text{sec}$$
Trying to find $dfrac{dD}{dt}$ by using formula $$V=pi r^2h$$
calculus
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
$endgroup$
I got this task on my calculus class and I got stuck at process of figuring it out
A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.
What I know:
Volume is unknown and constant
Rate of change of diameter is $$dfrac {dD}{dt} = 0,2text{cm}/text{sec}$$
Rate of change of height is $$dfrac {dh}{dt} = -0,4text{cm}/text{sec}$$
Trying to find $dfrac{dD}{dt}$ by using formula $$V=pi r^2h$$
calculus
calculus
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
edited Jan 17 at 9:36
Evgeniy
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
asked Jan 17 at 9:19
EvgeniyEvgeniy
134
134
add a comment |
add a comment |
1 Answer
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So, First we know that $$V=frac{pi D^2 h}{4} $$ where $D$ is the Diameter, and $h$ is the height.
Now by differentiating this with respect to time, we get,
$frac{dV}{dt} = frac {partial V}{partial D}frac{dD}{dt} + frac {partial V}{partial h}frac{dh}{dt}$
$;;;;;= frac{pi D h}{2} frac{dD}{dt} + frac{pi D^2}{4} frac{dh}{dt}$
Now, $frac {dV}{dt} = 0 ; ; frac{dD}{dt}=2; ; frac{dh}{dt} = -4; ;;; and ;;; h=10 $
(Note: All units are in mm)
Solving for $D$ we get, $D=10 mm$
Now coming back to the question we are needed to find $$frac {dA}{dt}$$
where $A= frac {pi D^2}{4}$ (Cross sectional area)
Taking derivative with respect to time, we get,
$frac {dA}{dt} = frac {pi D}{2} frac {dD}{dt}$
Plugging in values, we obtain,
$mathbf {frac {dA}{dt} = 10 pi ; ;mm^2/s}$
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
So, First we know that $$V=frac{pi D^2 h}{4} $$ where $D$ is the Diameter, and $h$ is the height.
Now by differentiating this with respect to time, we get,
$frac{dV}{dt} = frac {partial V}{partial D}frac{dD}{dt} + frac {partial V}{partial h}frac{dh}{dt}$
$;;;;;= frac{pi D h}{2} frac{dD}{dt} + frac{pi D^2}{4} frac{dh}{dt}$
Now, $frac {dV}{dt} = 0 ; ; frac{dD}{dt}=2; ; frac{dh}{dt} = -4; ;;; and ;;; h=10 $
(Note: All units are in mm)
Solving for $D$ we get, $D=10 mm$
Now coming back to the question we are needed to find $$frac {dA}{dt}$$
where $A= frac {pi D^2}{4}$ (Cross sectional area)
Taking derivative with respect to time, we get,
$frac {dA}{dt} = frac {pi D}{2} frac {dD}{dt}$
Plugging in values, we obtain,
$mathbf {frac {dA}{dt} = 10 pi ; ;mm^2/s}$
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
$endgroup$
add a comment |
$begingroup$
So, First we know that $$V=frac{pi D^2 h}{4} $$ where $D$ is the Diameter, and $h$ is the height.
Now by differentiating this with respect to time, we get,
$frac{dV}{dt} = frac {partial V}{partial D}frac{dD}{dt} + frac {partial V}{partial h}frac{dh}{dt}$
$;;;;;= frac{pi D h}{2} frac{dD}{dt} + frac{pi D^2}{4} frac{dh}{dt}$
Now, $frac {dV}{dt} = 0 ; ; frac{dD}{dt}=2; ; frac{dh}{dt} = -4; ;;; and ;;; h=10 $
(Note: All units are in mm)
Solving for $D$ we get, $D=10 mm$
Now coming back to the question we are needed to find $$frac {dA}{dt}$$
where $A= frac {pi D^2}{4}$ (Cross sectional area)
Taking derivative with respect to time, we get,
$frac {dA}{dt} = frac {pi D}{2} frac {dD}{dt}$
Plugging in values, we obtain,
$mathbf {frac {dA}{dt} = 10 pi ; ;mm^2/s}$
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
$endgroup$
add a comment |
$begingroup$
So, First we know that $$V=frac{pi D^2 h}{4} $$ where $D$ is the Diameter, and $h$ is the height.
Now by differentiating this with respect to time, we get,
$frac{dV}{dt} = frac {partial V}{partial D}frac{dD}{dt} + frac {partial V}{partial h}frac{dh}{dt}$
$;;;;;= frac{pi D h}{2} frac{dD}{dt} + frac{pi D^2}{4} frac{dh}{dt}$
Now, $frac {dV}{dt} = 0 ; ; frac{dD}{dt}=2; ; frac{dh}{dt} = -4; ;;; and ;;; h=10 $
(Note: All units are in mm)
Solving for $D$ we get, $D=10 mm$
Now coming back to the question we are needed to find $$frac {dA}{dt}$$
where $A= frac {pi D^2}{4}$ (Cross sectional area)
Taking derivative with respect to time, we get,
$frac {dA}{dt} = frac {pi D}{2} frac {dD}{dt}$
Plugging in values, we obtain,
$mathbf {frac {dA}{dt} = 10 pi ; ;mm^2/s}$
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
$endgroup$
So, First we know that $$V=frac{pi D^2 h}{4} $$ where $D$ is the Diameter, and $h$ is the height.
Now by differentiating this with respect to time, we get,
$frac{dV}{dt} = frac {partial V}{partial D}frac{dD}{dt} + frac {partial V}{partial h}frac{dh}{dt}$
$;;;;;= frac{pi D h}{2} frac{dD}{dt} + frac{pi D^2}{4} frac{dh}{dt}$
Now, $frac {dV}{dt} = 0 ; ; frac{dD}{dt}=2; ; frac{dh}{dt} = -4; ;;; and ;;; h=10 $
(Note: All units are in mm)
Solving for $D$ we get, $D=10 mm$
Now coming back to the question we are needed to find $$frac {dA}{dt}$$
where $A= frac {pi D^2}{4}$ (Cross sectional area)
Taking derivative with respect to time, we get,
$frac {dA}{dt} = frac {pi D}{2} frac {dD}{dt}$
Plugging in values, we obtain,
$mathbf {frac {dA}{dt} = 10 pi ; ;mm^2/s}$
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
answered Jan 17 at 10:08
The Jade EmperorThe Jade Emperor
908
908
add a comment |
add a comment |
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