Question

A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rate of 35 cm3/sec. The base radius of the tank is 1 meter and the height of the tank is 2.5 meters. When the depth of the water is 1.25 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing?\

Answer 1

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s