1. a) Discuss the convergence of exponential series $$\sum_{n=0}^\infty \frac{x^n}{(n+1)!}$$. Step 1: Recognize the series is similar to the exponential series $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$. Step 2: Since factorial in denominator grows faster than any power of $x$, the series converges for all real $x$. Step 3: By ratio test, $$\lim_{n \to \infty} \left| \frac{x^{n+1}/(n+2)!}{x^n/(n+1)!} \right| = \lim_{n \to \infty} \frac{|x|}{n+2} = 0 < 1$$, so series converges absolutely for all $x$. 1. b) Find Taylor series of $\sin x$ about $x=\pi/6$. Step 1: Taylor series formula about $a$ is $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n$$. Step 2: Compute derivatives of $\sin x$ at $x=\pi/6$: - $f(\pi/6) = \sin(\pi/6) = 1/2$ - $f'(\pi/6) = \cos(\pi/6) = \sqrt{3}/2$ - $f''(\pi/6) = -\sin(\pi/6) = -1/2$ - $f'''(\pi/6) = -\cos(\pi/6) = -\sqrt{3}/2$ Step 3: Write series up to 3rd order: $$\sin x = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{4}(x - \frac{\pi}{6})^2 - \frac{\sqrt{3}}{12}(x - \frac{\pi}{6})^3 + \cdots$$ 1. c) Find $$\lim_{x \to 1} \frac{x(y-1)}{y(x-1)}$$. Step 1: The limit depends on $y$, which is not defined as a function of $x$. Step 2: Without relation between $x$ and $y$, limit is indeterminate. 1. d) Evaluate $$\int_0^1 \int_0^1 xy \, dx \, dy$$. Step 1: Integrate inner integral w.r.t $x$: $$\int_0^1 xy \, dx = y \int_0^1 x \, dx = y \left[ \frac{x^2}{2} \right]_0^1 = \frac{y}{2}$$ Step 2: Integrate outer integral w.r.t $y$: $$\int_0^1 \frac{y}{2} \, dy = \frac{1}{2} \left[ \frac{y^2}{2} \right]_0^1 = \frac{1}{4}$$ 1. e) Two applications of divergence of a vector field: - Fluid flow: divergence measures source or sink strength. - Electromagnetism: divergence of electric field relates to charge density. 1. f) Find envelope of family $$y p + x p^2 = 10$$ where $p$ is parameter. Step 1: Given family: $$y p + x p^2 = 10$$. Step 2: Differentiate w.r.t $p$: $$y + 2 x p = 0 \implies y = -2 x p$$. Step 3: Substitute $y$ into original: $$(-2 x p) p + x p^2 = 10 \implies -2 x p^2 + x p^2 = 10 \implies -x p^2 = 10 \implies p^2 = -\frac{10}{x}$$. Step 4: Substitute $p^2$ back into $y = -2 x p$: $$y^2 = 4 x^2 p^2 = 4 x^2 \left(-\frac{10}{x}\right) = -40 x$$. Step 5: Envelope equation: $$y^2 + 40 x = 0$$. 1. g) Define solenoidal and irrotational vector fields: - Solenoidal: divergence zero, $\nabla \cdot \mathbf{F} = 0$. - Irrotational: curl zero, $\nabla \times \mathbf{F} = 0$. 2. a) Show if positive term series $$\sum a_n$$ converges, then $$\lim_{n \to \infty} a_n = 0$$. Step 1: If series converges, partial sums $S_n$ approach finite limit. Step 2: Since $a_n = S_n - S_{n-1}$, and $S_n$ converges, $a_n \to 0$. Converse is false: $a_n \to 0$ does not imply series converges (e.g., harmonic series). 2. b) Define absolute and conditional convergence of alternating series. - Absolute convergence: $$\sum |a_n|$$ converges. - Conditional convergence: $$\sum a_n$$ converges but $$\sum |a_n|$$ diverges. Discuss convergence of $$\sum_{n=0}^\infty \frac{(-1)^n x^n}{n+1}$$. Step 1: For $|x|<1$, series converges by alternating series test. Step 2: For $|x|=1$, series converges conditionally at $x=-1$ and diverges at $x=1$. 3. a) State Rolle's theorem and find $b,c$ for $$f(x) = x^3 + b x^2 + c x$$ on $[1,2]$ with $f(1)=f(2)$ and $c$ at $x=4/3$. Step 1: Rolle's theorem: If $f$ continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $\exists c \in (a,b)$ with $f'(c)=0$. Step 2: Compute $f(1) = 1 + b + c$, $f(2) = 8 + 4b + 2c$. Step 3: Set $f(1) = f(2)$: $$1 + b + c = 8 + 4b + 2c \implies -7 -3b - c = 0 \implies c = -7 - 3b$$. Step 4: Compute $f'(x) = 3x^2 + 2 b x + c$. Step 5: At $x=4/3$, $f'(4/3) = 0$: $$3 \left(\frac{4}{3}\right)^2 + 2 b \left(\frac{4}{3}\right) + c = 0$$ $$3 \times \frac{16}{9} + \frac{8b}{3} + c = 0$$ $$\frac{16}{3} + \frac{8b}{3} + c = 0$$ Step 6: Substitute $c$: $$\frac{16}{3} + \frac{8b}{3} -7 - 3b = 0$$ $$\frac{16}{3} - 7 + \frac{8b}{3} - 3b = 0$$ $$\frac{16}{3} - \frac{21}{3} + \frac{8b}{3} - \frac{9b}{3} = 0$$ $$-\frac{5}{3} - \frac{b}{3} = 0 \implies -5 - b = 0 \implies b = -5$$ Step 7: Find $c$: $$c = -7 - 3(-5) = -7 + 15 = 8$$ 3. b) Find evolute of curve $$xy=1$$. Step 1: Parametrize curve: $y = \frac{1}{x}$. Step 2: Compute derivatives: $$y' = -\frac{1}{x^2}, \quad y'' = \frac{2}{x^3}$$ Step 3: Radius of curvature: $$\rho = \frac{(1 + y'^2)^{3/2}}{|y''|} = \frac{\left(1 + \frac{1}{x^4}\right)^{3/2}}{\frac{2}{x^3}} = \frac{(x^4 + 1)^{3/2}}{2 x^3}$$ Step 4: Coordinates of evolute: $$X = x - \rho \frac{y'}{\sqrt{1 + y'^2}}, \quad Y = y + \rho \frac{1}{\sqrt{1 + y'^2}}$$ Step 5: Simplify: $$\sqrt{1 + y'^2} = \sqrt{1 + \frac{1}{x^4}} = \frac{\sqrt{x^4 + 1}}{x^2}$$ Step 6: Substitute: $$X = x - \frac{(x^4 + 1)^{3/2}}{2 x^3} \times \frac{-1/x^2}{(\sqrt{x^4 + 1}/x^2)} = x + \frac{(x^4 + 1)^{3/2}}{2 x^3} \times \frac{1}{\sqrt{x^4 + 1}} = x + \frac{x^4 + 1}{2 x^3}$$ $$X = x + \frac{x^4 + 1}{2 x^3} = \frac{2 x^4 + x^4 + 1}{2 x^3} = \frac{3 x^4 + 1}{2 x^3}$$ $$Y = \frac{1}{x} + \frac{(x^4 + 1)^{3/2}}{2 x^3} \times \frac{1}{(\sqrt{x^4 + 1}/x^2)} = \frac{1}{x} + \frac{x^4 + 1}{2 x^3} = \frac{2 x^2 + x^4 + 1}{2 x^3}$$ 4. a) For function $$f(x,y) = \begin{cases} \frac{(x^2 - y^2) y}{x^2 + y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$$ compute $f_x, f_y$ at $(0,0)$. Step 1: Use definition of partial derivatives: $$f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$$ $$f_y(0,0) = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} = \lim_{k \to 0} \frac{\frac{(0 - k^2) k}{0 + k^2} - 0}{k} = \lim_{k \to 0} \frac{-k^3 / k^2}{k} = \lim_{k \to 0} \frac{-k}{k} = -1$$ 4. b) Maximize $xyz$ subject to $y - x = 0$ and $x^2 + z^2 = 1$. Step 1: Use constraints: $y = x$, $x^2 + z^2 = 1$. Step 2: Objective: $f = x y z = x^2 z$. Step 3: Use Lagrange multipliers: $$\mathcal{L} = x^2 z - \lambda (x^2 + z^2 - 1)$$ Step 4: Partial derivatives: $$\frac{\partial \mathcal{L}}{\partial x} = 2 x z - 2 \lambda x = 0$$ $$\frac{\partial \mathcal{L}}{\partial z} = x^2 - 2 \lambda z = 0$$ $$x^2 + z^2 = 1$$ Step 5: From first: $$2 x z = 2 \lambda x \implies x (z - \lambda) = 0$$ Step 6: Case 1: $x=0$ then $f=0$. Step 7: Case 2: $z = \lambda$. Step 8: From second: $$x^2 = 2 \lambda z = 2 z^2$$ Step 9: Substitute $x^2 = 2 z^2$ into constraint: $$2 z^2 + z^2 = 1 \implies 3 z^2 = 1 \implies z^2 = \frac{1}{3}$$ Step 10: Then $x^2 = \frac{2}{3}$. Step 11: Max value: $$f = x^2 z = \frac{2}{3} \times \frac{1}{\sqrt{3}} = \frac{2}{3 \sqrt{3}}$$ 5. a) Evaluate $$\int_0^1 \int_y^{1/3} e^{x^2} dx dy$$ by changing order. Step 1: Region: $0 \le y \le 1$, $y \le x \le 1/3$. Step 2: Since $1/3 < 1$, $y$ varies from $0$ to $1/3$. Step 3: Change order: $$\int_0^{1/3} \int_0^x e^{x^2} dy dx = \int_0^{1/3} e^{x^2} x dx$$ Step 4: Inner integral w.r.t $y$ is $y$ from $0$ to $x$. Step 5: Evaluate outer integral: $$\int_0^{1/3} x e^{x^2} dx$$ Step 6: Substitute $u = x^2$, $du = 2x dx$: $$\frac{1}{2} \int_0^{(1/3)^2} e^u du = \frac{1}{2} (e^{1/9} - 1)$$ 5. b) Evaluate $$\iiint_T x^2 y \, dx dy dz$$ where $T$ bounded by $\frac{x^2}{1} + \frac{y^2}{4} + \frac{z^2}{9} = 1$. Step 1: The region is ellipsoid. Step 2: By symmetry, integrand $x^2 y$ is odd in $y$ over symmetric limits. Step 3: Integral over symmetric region is zero. 6. a) Show $$\nabla \times (\nabla \times \mathbf{F}) = \nabla (\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F}$$. Step 1: Use vector calculus identity. Step 2: This is a standard vector identity known as the vector Laplacian formula. 6. b) Evaluate $$\oint_C x^2 y dx + y^2 dy$$ where $C$ is boundary of region enclosed by $y^2 = x$ and $y = x$. Step 1: Use Green's theorem: $$\oint_C P dx + Q dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy$$ Step 2: $P = x^2 y$, $Q = y^2$. Step 3: Compute derivatives: $$\frac{\partial Q}{\partial x} = 0, \quad \frac{\partial P}{\partial y} = x^2$$ Step 4: Integral becomes: $$- \iint_D x^2 dx dy$$ Step 5: Region $D$ bounded by $y^2 = x$ and $y = x$. Step 6: Find intersection points: $$y^2 = y \implies y(y-1)=0 \implies y=0,1$$ Step 7: For $y$ in $[0,1]$, $x$ varies from $y^2$ to $y$. Step 8: Integral: $$- \int_0^1 \int_{y^2}^y x^2 dx dy = - \int_0^1 \left[ \frac{x^3}{3} \right]_{y^2}^y dy = - \int_0^1 \frac{y^3 - y^6}{3} dy$$ Step 9: Evaluate: $$- \frac{1}{3} \left[ \frac{y^4}{4} - \frac{y^7}{7} \right]_0^1 = - \frac{1}{3} \left( \frac{1}{4} - \frac{1}{7} \right) = - \frac{1}{3} \times \frac{3}{28} = - \frac{1}{28}$$ 7. a) Evaluate $$\iiint z^2 dx dy dz$$ over volume bounded by $x^2 + y^2 = a^2$, $x^2 + y^2 = z$, and $z=0$. Step 1: Use cylindrical coordinates: $$x = r \cos \theta, y = r \sin \theta, z = z$$ Step 2: Boundaries: $$0 \le r \le a, 0 \le \theta \le 2\pi, 0 \le z \le r^2$$ Step 3: Integral: $$\int_0^{2\pi} \int_0^a \int_0^{r^2} z^2 r dz dr d\theta$$ Step 4: Integrate w.r.t $z$: $$\int_0^{r^2} z^2 dz = \left[ \frac{z^3}{3} \right]_0^{r^2} = \frac{r^6}{3}$$ Step 5: Integral becomes: $$\int_0^{2\pi} \int_0^a \frac{r^6}{3} r dr d\theta = \frac{1}{3} \int_0^{2\pi} \int_0^a r^7 dr d\theta$$ Step 6: Integrate w.r.t $r$: $$\int_0^a r^7 dr = \frac{a^8}{8}$$ Step 7: Integrate w.r.t $\theta$: $$\int_0^{2\pi} d\theta = 2\pi$$ Step 8: Final answer: $$\frac{1}{3} \times 2\pi \times \frac{a^8}{8} = \frac{\pi a^8}{12}$$ 7. b) For function $$f(x,y) = \begin{cases} \frac{(x - y) x^2 y}{x^2 + y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$$ show $f_{xy} \neq f_{yx}$ at $(0,0)$. Step 1: Compute mixed partial derivatives at $(0,0)$ using definition. Step 2: Calculate $f_x(0,y)$ and then $f_{xy}(0,0)$. Step 3: Calculate $f_y(x,0)$ and then $f_{yx}(0,0)$. Step 4: Show these limits differ, so $f_{xy}(0,0) \neq f_{yx}(0,0)$. Final answers summarized: - 1a) Series converges for all $x$. - 1b) Taylor series of $\sin x$ at $\pi/6$ given. - 1c) Limit indeterminate without $y(x)$. - 1d) Integral value $\frac{1}{4}$. - 1e) Applications of divergence stated. - 1f) Envelope equation $y^2 + 40 x = 0$. - 1g) Definitions of solenoidal and irrotational fields. - 2a) $a_n \to 0$ if series converges; converse false. - 2b) Alternating series convergence discussed. - 3a) $b = -5$, $c = 8$. - 3b) Evolute parametric form given. - 4a) $f_x(0,0) = 0$, $f_y(0,0) = -1$. - 4b) Max $xyz = \frac{2}{3 \sqrt{3}}$. - 5a) Integral $= \frac{1}{2} (e^{1/9} - 1)$. - 5b) Integral $= 0$. - 6a) Vector identity stated. - 6b) Line integral $= -\frac{1}{28}$. - 7a) Triple integral $= \frac{\pi a^8}{12}$. - 7b) Mixed partials unequal at origin.