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Yakvenalex [24]
3 years ago
15

A cylinder shaped container has a radius of 30 cm and a height of 100 cm. A glass sphere with a redius of 12 cm is placed inside

the cylinder shaped container and then the container is completely filled with water. What is the approximate volume of water contained in the cylinder? Use 3.14 as an approximate for pi.
7,235 cm3

28,260 cm3

204,715 cm3

275,365 cm3
Mathematics
1 answer:
Artemon [7]3 years ago
4 0
Vcylinder-Vsphere=water


Vcylinder=hpir^2
Vsphere=(4/3)pir^3


given

cylinderradius=30 and h=100
Vcylinder=100pi30^2=100pi900=90000pi
let's leave it in terms of pi for more exactness

sphereradius=12
Vsphere=(4/3)pi12^3=2304pi


cylinder-sphere=90000pi-2304pi=87696pi
using pi=3.14
water=275365.44 cubic cm

last option is corrrect
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(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

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Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

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But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
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Solve inside of the parentheses first.

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