MATH SOLVE

8 months ago

Q:
# 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.

Accepted Solution

A:

Answer:28.27 cm/sStep-by-step explanation:Though Process: The punch glass (call it bowl to have a shape in mind) is in the shape of a hemispherethe radius [tex]r=5cm[/tex] Punch is being poured into the bowlThe height at which the punch is increasing in the bowl is [tex]\frac{dh}{dt} = 1.5[/tex]the exposed area is a circle, (since the bowl is a hemisphere)the radius of this circle can be written as [tex]'a'[/tex]what is being asked is the rate of change of the exposed area when the height [tex]h = 2 cm[/tex] the rate of change of exposed area can be written as [tex]\frac{dA}{dt}[/tex]. since the exposed area is changing with respect to the height of punch. We can use the chain rule: [tex]\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}[/tex]and since [tex]A = \pi a^2[/tex] the chain rule above can simplified to [tex]\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}[/tex] -- we can call this Eq(1)Solution: the area of the exposed circle is[tex]A =\pi a^2 [/tex]the rate of change of this area can be, (using chain rule)[tex]\frac{dA}{dt} = 2 \pi a \frac{da}{dt}[/tex] we can call this Eq(2)what we are really concerned about is how [tex]a[/tex] changes as the punch is being poured into the bowl i.e [tex]\frac{da}{dh}[/tex]So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:[tex]r = \frac{a^2 + h^2}{2h}[/tex]and rearrage the formula so that a is the subject:[tex]a^2 = 2rh - h^2[/tex]now we can derivate a with respect to h to get [tex]\frac{da}{dh}[/tex][tex]2a \frac{da}{dh} = 2r - 2h[/tex]simplify[tex]\frac{da}{dh} = \frac{r-h}{a}[/tex]we can put this in Eq(1) in place of [tex]\frac{da}{dh}[/tex][tex]\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}[/tex] and since we know [tex]\frac{dh}{dt} = 1.5[/tex][tex]\frac{da}{dt} = \frac{(r-h)(1.5)}{a} [/tex]and now we use substitute this [tex]\frac{da}{dt}[/tex]. in Eq(2)[tex]\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}[/tex]simplify,[tex]\frac{dA}{dt} = 3 \pi (r-h)[/tex]This is the rate of change of area, this is being asked in the quesiton! Finally, we can put our known values:[tex]r = 5cm[/tex][tex]h = 2cm[/tex] from the question[tex]\frac{dA}{dt} = 3 \pi (5-2)[/tex][tex]\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s[/tex]