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Vlad1618 [11]
3 years ago
9

How do you calculate the derivative of y with respect to x on the equation x^3y+3xy^3=x+y?

Mathematics
1 answer:
lesya692 [45]3 years ago
3 0
\bf x^3y+3xy^3=x+y\\\\
-----------------------------\\\\
\left[ 3x^2y+x^3\frac{dy}{dx} \right]+3\left[ 1\cdot y^3+x\cdot 3y^2\frac{dy}{dx} \right]=1+\frac{dy}{dx}
\\\\\\
3x^2y+x^3\frac{dy}{dx}+3y^3+9xy^2\frac{dy}{dx}-\frac{dy}{dx}=1\impliedby \textit{common factor}
\\\\\\
\cfrac{dy}{dx}(x^3+9xy^2-1)+3x^2y+3y^3=1
\\\\\\
\cfrac{dy}{dx}(x^3+9xy^2-1)=1-3x^2y-3y^3
\\\\\\
\cfrac{dy}{dx}=\cfrac{1-3x^2y-3y^3}{x^3+9xy^2-1}
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Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,
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If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

V_{flask}=V_{sphere}+V_{cylinder}.

Use following formulas to determine volumes of sphere and cylinder:

V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

  • V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;
  • V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;
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Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.

Then

\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.

Answer 2: correct choice is D.


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