Question

A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the change in height of the punch is 1,5 cm/sec, at what rate is the exposed area of the punch changing when the height of the punch is 2 cm.

Answer 1

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

• The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
• the radius $r=5cm$
• Punch is being poured into the bowl
• The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
• the exposed area is a circle, (since the bowl is a hemisphere)
• the radius of this circle can be written as $'a'$

• what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
• the rate of change of exposed area can be written as $\frac{dA}{dt}$.
• since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

• and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$. in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$