Bacteria tend to grow very fast in a Petri dish at first because of unlimited food and then begin to die out due

Mathematics

1 answer:

Assoli18 [71]3 years ago

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Answer:

6. Bacteria tend to grow very fast in a petri dish at first because of unlimited food and then begin to die out due

to competition. In a certain culture, the number of bacteria is givenby N(f) =1f +92t+.625' where /

represents the hours since 625 bac,teriawere introduced to the Petri dish. Determine the *={TlgX9"t

of bacteria that occur in the dish.

Step-by-step explanation:

t= -92/-2(2) =23

N (23) power 2 +92(23) + 625 =1628

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Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

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Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

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Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$