Answer:
See the step by step solution
Step-by-step explanation:
Consider a more general case of the form
. We want to express it as a sine term only by using the following formula ![\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%2B%5Cbeta%29%20%3D%20%5Csin%28%5Calpha%29%5Ccos%28%5Cbeta%29%2B%5Csin%28%5Cbeta%29%5Ccos%28%5Calpha%29)
The trick consists on find a number
such that
. But, note that since
it is most likely that
. Then, we will use the following equations:
![\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B](https://tex.z-dn.net/?f=%5Ccos%28%5Cbeta%29%20%3D%5Cfrac%7Ba%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DA%2C%5Csin%28%5Cbeta%29%20%3D%5Cfrac%7Bb%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DB)
Then, problem turns out to be
,
where ![\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7BB%7D%7BA%7D%29%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bb%7D%7Ba%7D%29.%20So%2C%20note%20that%20the%20frequency%20is%20the%20same.%20%3C%2Fp%3E%3Cp%3EThen%2C%20%3C%2Fp%3E%3Cp%3Ea%29%20Taking%20a%3D4%20and%20b%3D%2015%20we%20have%20that%204%20sin%202%CF%80t%20%2B%2015%20cos%202%CF%80t%3D%20%5Btex%5D%5Csqrt%5B%5D%7B15%5E2%2B4%5E2%7D%5Csin%282%5Cpi%20t%20%2B%20%5Ctan%5E%7B-1%7D%5Cfrac%7B15%7D%7B4%7D%29)
b) The amplitude is
. Since
the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.
c) The function's graph is attached
500 • .03
the answer of that times two. then there is the answer
The dimensions of the can are 1.457 inches and 2.913 inches that will give the most volume
Step-by-step explanation:
Let us revise the rule of surface area and volume of a cylinder
- S.A = 2π r h + 2π r²
- V = π r² h
Forty square inches of material is available to make a cylindrical; can of tuna and water, we need to find the dimensions of the can that will give the most volume
∵ S.A = 40 inches²
∵ S.A = 2π r h + 2π r²
∴ 2π r h + 2π r² = 40
Let us use this rule to find h in terms of r
- Subtract 2π r² from both sides
∵ 2π r h = 40 - 2 π r²
- Divide both sides by 2π r
∴ ![h=\frac{40-2\pi r^{2}}{2\pi r}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B40-2%5Cpi%20r%5E%7B2%7D%7D%7B2%5Cpi%20r%7D)
∴ ![h=\frac{40}{2\pi r}-\frac{2\pi r^{2}}{2\pi r}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B40%7D%7B2%5Cpi%20r%7D-%5Cfrac%7B2%5Cpi%20r%5E%7B2%7D%7D%7B2%5Cpi%20r%7D)
∴ ![h=\frac{20}{\pi r}-r](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B20%7D%7B%5Cpi%20r%7D-r)
∵ V = π r² h
- Substitute h by its value above
∴ ![V=\pi r^{2}(\frac{20}{\pi r}-r)](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E%7B2%7D%28%5Cfrac%7B20%7D%7B%5Cpi%20r%7D-r%29)
∴ V = 20 r - π r³
To find the most volume differentiate it with respect to r and equate it by 0 to find the value of r
∵
= 20 - 3π r²
∵
= 0
∴ 20 - 3π r² = 0
- Add 3π r² to both sides
∴ 20 = 3π r²
- Divide both sides by 3π
∴ r² = 2.122
- Take √ for both sides
∴ r = 1.457 inches
To find h substitute the value of r in the expression of h
∵ ![h=\frac{20}{\pi r}-r](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B20%7D%7B%5Cpi%20r%7D-r)
∴ ![h=\frac{20}{\pi (1.457)}-(1.457)](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B20%7D%7B%5Cpi%20%281.457%29%7D-%281.457%29)
∴ h = 2.913 inches
The dimensions of the can are 1.457 inches and 2.913 inches that will give the most volume
Learn more:
You can learn more about volume of solids in brainly.com/question/6443737
#LearnwithBrainly