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Maurinko [17]
4 years ago
5

See attached image. For part b, my answer is

%5Csqrt%7B17%7Dt%20%7D%7B5%7D%2B17" id="TexFormula1" title="h(t)=\frac{t^2}{100}-\frac{\sqrt{17}t }{5}+17" alt="h(t)=\frac{t^2}{100}-\frac{\sqrt{17}t }{5}+17" align="absmiddle" class="latex-formula"> , 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

Mathematics
1 answer:
ArbitrLikvidat [17]4 years ago
5 0

Answer:

a) shown

b) h = [sqrt(17) - (5/2)t]²

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

Step-by-step 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

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