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

The rate at which the area of the punch change = 165 cm²/sec

Step-by-step explanation:

Using chain rule,

dv/dt = dh/dt × dv/dr ............................. equation 1

Where dA/dt = rate of change of the exposed area, dh/dt = rate of change of height, dA/dh = differentiation of the area of the glass with respect to height

The glass is cylindrical in shape,

∴ surface area (A) = πr² + 2πrh ........................... equation 2

Differentiating equation1 with respect to h

dA/dh = πr² + 2πr.

dh/dt = 1.5 cm/sec, r = 5cm, π = 3.143

Substituting these values into equation 1,

dA/dt = 1.5 × {(3.143×5²) +(2×3.143×5)}

dA/dt = 1.5 × (78.575 + 31.43)

dA/dt = 1.5 × 110.005

dA/dt = 165.0075

∴ dA/dt ≈ 165 cm²/sec

The rate at which the area of the punch change = 165 cm²/sec

Answer 2

28.27 cm2/s

Step-by-step explanation:

Let's h be the height of the punch. We can calculate the exposed area of the punch in term of h.

The radius of the exposed surface, the glass radius, and the distance from the exposed surface to the top glass surface forms a right triangle.

glass radius is 5 cm. Distance from the punch surface to top of glass surface is 5 - h. The the exposed radius of the punch is

`r^2 = 5^2 - (5 - h)^2 = 25 - 25 +10h - h^2 = -h^2 + 10h`

Therefore the area of the exposed surface is

`A = \pi r^2 = \pi h(10 - h)`

Using chain rule we can calculate the rate of change for area

`(dA)/(dt) = (dA)/(dh)(dh)/(dt) = d(10h\pi - \pi h^2)/(dh) 1.5`

`1.5\pi(10 - 2h)`

`3\pi(5 - h)`

Here we can substitute h = 2cm

`(dA)/(dt) = 3\pi(5 - 2) = 9\pi = 28.27 cm^2/s`