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See attached image. For part b, my answer is
h(t)=(t^2)/(100)-(√(17)t )/(5)+17 , and for part c I think the coffee pot is empty at t=41.231, although I'm not sure about these answers.

Edit: I solved this one, no need to answer

See attached image. For part b, my answer is h(t)=(t^2)/(100)-(√(17)t )/(5)+17 , and-example-1

1 Answer

5 votes

Answer:

a) shown

b) h = [sqrt(17) - (5/2)t]²

c) t = 2sqrt(17)/5 seconds

Explanation:

V = pi × r² × h

V = pi × 5² × h

V = 25pi × h

a) dV/dt = dV/dh × dh/dt

-5pi × sqrt(h) = 25pi × dh/dt

dh/dt = -sqrt(h)/5

b) 1/sqrt(h) .dh = -5. dt

2sqrt(h) = -5t + c

t = 0, h = 17

2sqrt(17) = 0 + c

c = 2sqrt(17)

2sqrt(h) = -5t + 2sqrt(17)

sqrt(h) = [2sqrt(17) - 5t] ÷ 2

sqrt(h) = sqrt(17) - (5/2)t

Square both sides

h = [sqrt(17) - (5/2)t]²

c) empty: h = 0

0 = [sqrt(17) - (5/2)t]²

sqrt(17) - (5/2)t = 0

(5/2)t = sqrt(17)

t = 2sqrt(17)/5

t = 1.64924225 seconds

sqrt: square root

User Zizazorro
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