Homework problems

cal quw wus69

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Aug 31, 2010
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1.A spherical balloon with radius r inches has volume defined by the function below. Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 2 inches. (Give the answer in terms of ? and r.)

V(r)=4/3(pi)r(to the third)... Already tried V(r)=4/3(pi)(r+2)(to the third).. didnt work

2.Find the domain of the function. (If you need to use - or , enter -INFINITY or INFINITY.)

G(u)=square root of (u) + square root of(1-u)... already tried (1,-infinity)

sorry about the wierd functions but im new to this form and dont know how to write equations in it yet.. ahh Calculus is a bitch and if someone could help i would really appreciate it
 
2) g(u) = u+1u\displaystyle 2) \ g(u) \ = \ \sqrt u+\sqrt{1-u}

Now. in real number land the  square root of anything must be  0.\displaystyle Now. \ in \ real \ number \ land \ the \ \ square \ root \ of \ anything \ must \ be \ \ge \ 0.

Hence, u  0 and 1u  0,      u  0 and u  1, ergo\displaystyle Hence, \ u \ \ge \ 0 \ and \ 1-u \ \ge \ 0, \ \implies \ u \ \ge \ 0 \ and \ u \ \le \ 1, \ ergo

the domain of g(u) is [0,1], QED\displaystyle the \ domain \ of \ g(u) \ is \ [0,1], \ QED

Note: We have to take each terms intersection so the whole equation is satisfied,\displaystyle Note: \ We \ have \ to \ take \ each \ term's \ intersection \ so \ the \ whole \ equation \ is \ satisfied,

or (,1] intersection [0,)\displaystyle or \ (-\infty,1] \ intersection \ [0,\infty)
 
cal quw wus69 said:
1.A spherical balloon with radius r inches has volume defined by the function below. Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 2 inches. (Give the answer in terms of ? and r.)

The amount of air required to inflate the balloon from 43πr3\displaystyle \frac{4}{3}{\pi}r^{3} to 43π(r+2)3\displaystyle \frac{4}{3}{\pi}(r+2)^{3}

Would be the difference between the two volumes.

43π(r+2)3final amount43πr3initial amount=8π(3r2+6r+4)3amount added\displaystyle \underbrace{\frac{4}{3}{\pi}(r+2)^{3}}_{\text{final amount}}-\overbrace{\frac{4}{3}{\pi}r^{3}}^{\text{initial amount}}=\underbrace{\frac{8\pi(3r^{2}+6r+4)}{3}}_{\text{amount added}}
 
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