a) Find the rate of change of volume of a sphere with respect to its surface area when radius is 7 cm. [2] b) Integrate: dx5 3 x [3] c) Solve : (1+x^2)dy (1+y^2)dx = 0 [3]

a) Find the rate of change of volume of a sphere with respect to its surface area when radius is 7 cm.

Step 1: Recall the formulas for volume and surface area of a sphere

• Volume of a sphere: $V=\frac{4}{3}\pi r^3$
• Surface area of a sphere: $S=4\pi r^2$

Step 2: Find the rate of change of volume with respect to radius Differentiate $V$ with respect to $r$: $$\frac{dV}{dr}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)=4\pi r^2$$

Step 3: Find the rate of change of surface area with respect to radius Differentiate $S$ with respect to $r$: $$\frac{dS}{dr}=\frac{d}{dr}\left(4\pi r^2\right)=8\pi r$$

Step 4: Use the chain rule to find $\frac{dV}{dS}$ We want $\frac{dV}{dS}$, which can be found using the chain rule: $$\frac{dV}{dS}=\frac{dV/dr}{dS/dr}=\frac{4\pi r^2}{8\pi r}=\frac{r}{2}$$

Step 5: Substitute $r=7$ cm $$\frac{dV}{dS}{|}_{r=7}=\frac{7}{2}=3.5$$

Final Answer: The rate of change of volume with respect to surface area when the radius is 7 cm is 3.5 cm.

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b) Integrate: $\int \frac{dx}{5-3\cos x}$

Step 1: Use the substitution $t=\tan \frac{x}{2}$ Recall the Weierstrass substitution: $$\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2}{1+t^2}dt$$

Step 2: Rewrite the integral in terms of $t$ $$\int \frac{dx}{5-3\cos x}=\int \frac{\frac{2}{1+t^2}dt}{5-3\left(\frac{1-t^2}{1+t^2}\right)}$$

Step 3: Simplify the denominator $$5-3\left(\frac{1-t^2}{1+t^2}\right)=\frac{5(1+t^2)-3(1-t^2)}{1+t^2}=\frac{5+5t^2-3+3t^2}{1+t^2}=\frac{2+8t^2}{1+t^2}$$

Step 4: Rewrite the integral $$\int \frac{2}{2+8t^2}dt=\int \frac{2}{2(1+4t^2)}dt=\int \frac{1}{1+4t^2}dt$$

Step 5: Factor out constants and integrate $$\int \frac{1}{1+4t^2}dt=\frac{1}{4}\int \frac{1}{{\left(\frac{1}{2}\right)}^2+t^2}dt$$ Let $u=2t$, then $du=2dt$ and $dt=\frac{du}{2}$: $$\frac{1}{4}\int \frac{1}{1+u^2}\cdot \frac{du}{2}=\frac{1}{8}\int \frac{1}{1+u^2}du=\frac{1}{8}{\tan }^{-1}u+C$$ Substitute back $u=2t$: $$\frac{1}{8}{\tan }^{-1}(2t)+C$$

Step 6: Substitute back $t=\tan \frac{x}{2}$ $$\frac{1}{8}{\tan }^{-1}\left(2\tan \frac{x}{2}\right)+C$$

Final Answer: $$\boxed{\frac{1}{8}{\tan }^{-1}\left(2\tan \frac{x}{2}\right)+C}$$

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c) Solve: $(1+x^2)dy-(1+y^2)dx=0$

Step 1: Rewrite the equation in differential form $$(1+x^2)dy=(1+y^2)dx$$ $$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$$

Step 2: Separate variables $$\frac{dy}{1+y^2}=\frac{dx}{1+x^2}$$

Step 3: Integrate both sides $$\int \frac{dy}{1+y^2}=\int \frac{dx}{1+x^2}$$ $${\tan }^{-1}y={\tan }^{-1}x+C$$

Step 4: Solve for $y$ $${\tan }^{-1}y-{\tan }^{-1}x=C$$ Using the identity for ${\tan }^{-1}a-{\tan }^{-1}b$: $${\tan }^{-1}\left(\frac{y-x}{1+xy}\right)=C$$ $$\frac{y-x}{1+xy}=\tan C=k(\text{where }k\text{ is a constant})$$

Step 5: Solve for $y$ $$y-x=k(1+xy)$$ $$y-kxy=x+k$$ $$y(1-kx)=x+k$$ $$y=\frac{x+k}{1-kx}$$

Final Answer: The general solution is: $$\boxed{y=\frac{x+C}{1-Cx}}$$ where $C$ is an arbitrary constant.