a) Water is poured into a right circular cylinder of radius 8cm at the rate of 18cu.cm/min. Prove that the rate which the level of water is rising in the cylinder is 932 cm/min . [2] b) Evaluate :…

a) Rate of rise of water level in a cylinder

Given:

• Radius of cylinder, $r=8$ cm
• Rate of pouring water, $\frac{dV}{dt}=18$ cm³/min

To prove: The rate at which the water level rises is $\frac{9}{32\pi }$ cm/min.

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Step 1: Volume of a cylinder The volume $V$ of a right circular cylinder is given by: $$V=\pi r^{2}h$$ where $h$ is the height (or level) of water.

Step 2: Differentiate with respect to time Differentiate both sides with respect to time $t$: $$\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$$ Here, $\frac{dh}{dt}$ is the rate at which the water level rises.

Step 3: Substitute known values Given $\frac{dV}{dt}=18$ cm³/min and $r=8$ cm: $$18=\pi (8{)}^2\frac{dh}{dt}$$ $$18=64\pi \frac{dh}{dt}$$

Step 4: Solve for $\frac{dh}{dt}$ $$\frac{dh}{dt}=\frac{18}{64\pi }$$ Simplify the fraction: $$\frac{dh}{dt}=\frac{9}{32\pi }\text{ cm/min}$$

Conclusion: The rate at which the water level rises is $\frac{9}{32\pi }$ cm/min.

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b) Evaluate the integral:

$$\int \frac{x^2-1}{x^4+x^2+1}dx$$

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Step 1: Factor the denominator The denominator is $x^4+x^2+1$. Let $y=x^2$: $$y^2+y+1$$ This does not factor easily, so we use a substitution.

Step 2: Rewrite the integrand Notice that the numerator is $x^2-1$, and the denominator is $x^4+x^2+1$. Let’s attempt to express the numerator as a derivative of the denominator.

Let $u=x^4+x^2+1$. Then: $$\frac{du}{dx}=4x^3+2x$$ This does not match the numerator, so we try a different approach.

Step 3: Partial fractions or substitution Let’s rewrite the integrand: $$\frac{x^2-1}{x^4+x^2+1}=\frac{x^2+1-2}{x^4+x^2+1}=\frac{x^2+1}{x^4+x^2+1}-\frac{2}{x^4+x^2+1}$$

Step 4: Simplify the first term $$\frac{x^2+1}{x^4+x^2+1}=\frac{1}{x^2}\cdot \frac{x^2+1}{x^2+1+\frac{1}{x^2}}$$ This seems complicated. Instead, let’s use the substitution $x=\tan \theta$, but it may not simplify easily.

Alternative approach: Divide numerator and denominator by $x^2$ $$\frac{x^2-1}{x^4+x^2+1}=\frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}$$ Let $z=x-\frac{1}{x}$. Then: $$z^2=x^2-2+\frac{1}{x^2}⟹x^2+\frac{1}{x^2}=z^2+2$$ The denominator becomes: $$x^2+1+\frac{1}{x^2}=(x^2+\frac{1}{x^2})+1=z^2+3$$ The numerator is: $$1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$$ This does not directly match $z$, so we adjust.

Step 5: Rewrite the integrand Notice that: $$\frac{x^2-1}{x^4+x^2+1}=\frac{1}{2}\cdot \frac{2x^2-2}{x^4+x^2+1}$$ Let’s split the fraction: $$\frac{2x^2}{x^4+x^2+1}-\frac{2}{x^4+x^2+1}$$

Step 6: Simplify the first term $$\frac{2x^2}{x^4+x^2+1}=\frac{2}{x^2+1+\frac{1}{x^2}}$$ Let $t=x-\frac{1}{x}$, then $dt=\left(1+\frac{1}{x^2}\right)dx$. This seems messy.

Step 7: Use substitution $u=x^2$ Let $u=x^2$, $du=2xdx$. The integral becomes: $$\int \frac{x^2-1}{x^4+x^2+1}dx=\int \frac{u-1}{u^2+u+1}\cdot \frac{du}{2\sqrt{u}}$$ This is not straightforward.

Step 8: Correct substitution Let’s try $u=x-\frac{1}{x}$. Then: $$du=\left(1+\frac{1}{x^2}\right)dx⟹dx=\frac{du}{1+\frac{1}{x^2}}$$ This is not directly helpful.

Step 9: Factor the denominator The denominator $x^4+x^2+1$ can be factored as: $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ Now, perform partial fraction decomposition: $$\frac{x^2-1}{(x^2+x+1)(x^2-x+1)}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}$$ Multiply both sides by the denominator: $$x^2-1=(Ax+B)(x^2-x+1)+(Cx+D)(x^2+x+1)$$ Expand and equate coefficients: $$x^2-1=A{x}^3-A{x}^2+Ax+B{x}^2-Bx+B+C{x}^3+C{x}^2+Cx+D{x}^2+Dx+D$$ Combine like terms: $$x^2-1=(A+C)x^3+(-A+B+C+D)x^2+(A-B+C+D)x+(B+D)$$ Equate coefficients:

1. $A+C=0$
2. $-A+B+C+D=1$
3. $A-B+C+D=0$
4. $B+D=-1$

Solve the system: From (1): $C=-A$. Substitute $C=-A$ into (2) and (3): $$-A+B-A+D=1⟹-2A+B+D=1$$ $$A-B-A+D=0⟹-B+D=0⟹D=B$$ From (4): $B+D=-1⟹2B=-1⟹B=-\frac{1}{2}$, $D=-\frac{1}{2}$. From (2): $-2A-\frac{1}{2}-\frac{1}{2}=1⟹-2A-1=1⟹-2A=2⟹A=-1$. From (1): $C=1$.

Thus: $$\frac{x^2-1}{(x^2+x+1)(x^2-x+1)}=\frac{-x-\frac{1}{2}}{x^2+x+1}+\frac{x-\frac{1}{2}}{x^2-x+1}$$

Step 10: Integrate each term $$\int \frac{-x-\frac{1}{2}}{x^2+x+1}dx+\int \frac{x-\frac{1}{2}}{x^2-x+1}dx$$

For the first integral: $$\int \frac{-x-\frac{1}{2}}{x^2+x+1}dx=-\frac{1}{2}\int \frac{2x+1}{x^2+x+1}dx-\frac{1}{2}\int \frac{dx}{x^2+x+1}$$ The first part is straightforward: $$-\frac{1}{2}\ln |x^2+x+1|$$ For the second part, complete the square: $$x^2+x+1={\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}$$ Thus: $$\int \frac{dx}{x^2+x+1}=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)$$

For the second integral: $$\int \frac{x-\frac{1}{2}}{x^2-x+1}dx=\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx+\frac{1}{2}\int \frac{dx}{x^2-x+1}$$ The first part is: $$\frac{1}{2}\ln |x^2-x+1|$$ For the second part, complete the square: $$x^2-x+1={\left(x-\frac{1}{2}\right)}^2+\frac{3}{4}$$ Thus: $$\int \frac{dx}{x^2-x+1}=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$$

Step 11: Combine results $$\int \frac{x^2-1}{x^4+x^2+1}dx=-\frac{1}{2}\ln |x^2+x+1|-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{1}{2}\ln |x^2-x+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$$ Simplify: $$=\frac{1}{2}\ln \left|\frac{x^2-x+1}{x^2+x+1}\right|+\frac{1}{\sqrt{3}}\left[{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]+C$$

Final Answer: $$\boxed{\frac{1}{2}\ln \left|\frac{x^2-x+1}{x^2+x+1}\right|+\frac{1}{\sqrt{3}}\left[{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]+C}$$

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c) Differential equation and polynomial solution

Given: $$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$$

Question: Is the solution a polynomial? Give reason.

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Step 1: Rewrite the differential equation $$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$$ This is a homogeneous differential equation. Let’s assume a solution of the form $y=vx$, where $v$ is a function of $x$.

Step 2: Substitute $y=vx$ $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$ Substitute into the original equation: $$v+x\frac{dv}{dx}=\frac{x^2+(vx{)}^2}{2x(vx)}=\frac{x^2+v^2{x}^2}{2v{x}^2}=\frac{1+v^2}{2v}$$ Simplify: $$v+x\frac{dv}{dx}=\frac{1+v^2}{2v}$$ Rearrange: $$x\frac{dv}{dx}=\frac{1+v^2}{2v}-v=\frac{1+v^2-2v^2}{2v}=\frac{1-v^2}{2v}$$ Separate variables: $$\frac{2v}{1-v^2}dv=\frac{1}{x}dx$$

Step 3: Integrate both sides $$\int \frac{2v}{1-v^2}dv=\int \frac{1}{x}dx$$ The left integral is: $$-\ln |1-v^2|=\ln |x|+C$$ Exponentiate both sides: $$|1-v^2|=e^{-\ln |x|+C}=\frac{e^C}{|x|}=\frac{K}{|x|}$$ where $K=e^C$. Thus: $$1-v^2=\frac{K}{x}$$ Substitute $v=\frac{y}{x}$: $$1-{\left(\frac{y}{x}\right)}^2=\frac{K}{x}$$ Multiply through by $x^2$: $$x^2-y^2=Kx$$ Rearrange: $$x^2-Kx-y^2=0$$

Step 4: Check if the solution is a polynomial The general solution is: $$x^2-y^2=Kx$$ This can be rewritten as: $$y^2=x^2-Kx$$ Taking the square root: $$y=\pm \sqrt{x^2-Kx}$$ The solution involves a square root, which is not a polynomial.

Conclusion: The solution is not a polynomial because it involves a square root term.

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Summary of Answers:

• (a) The rate of rise of water level is $\frac{9}{32\pi }$ cm/min.
• (b) The integral evaluates to: $$\boxed{\frac{1}{2}\ln \left|\frac{x^2-x+1}{x^2+x+1}\right|+\frac{1}{\sqrt{3}}\left[{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]+C}$$
• (c) The solution is not a polynomial because it involves a square root term.