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zubka84 [21]
4 years ago
14

Sketch the curve. r = 4 + 2 cos(θ) webassign plot webassign plot webassign plot webassign plot correct: your answer is correct.

find the area that it encloses.
Mathematics
1 answer:
Nitella [24]4 years ago
4 0
Make a table with the angle theta as independent variable and the radius r as dependent variable:

theta  radius = 4+2cos theta    radius
------- -----------------------------------------
   0      4+2                                6
  pi/6   4+2cos pi/6 =                 4+2(sqrt(3)/2

Perhaps you have already plotted this using webassign (but remember that you have not shared an illustration here).  (Please don't type "webassign plot" repeatedly, as it accomplishes nothing.)

Generally, when one wishes to find the area of a region defined by polar functions (as is the case here), one first determines suitable limits of integration from the finished curve and checks them through actual integration.  

Which formula should you use to find the area:  Look up "areas in polar coordinates," as I did.  The formula is as follows:


Enclosed area = Integral from alpha to beta of  (1/2)r^2 d(theta).  Note that the initial radius here is 6 (since r = 4 plus 2 cos theta is 4+2 when theta = 0). 
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The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

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