Calculus & Geometry Cheatsheet

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Q3) Integration Problems a) Evaluate $\int_0^1 \frac{x^2}{(1+x^2)^{3/2}} dx$ Substitution: Let $x = \tan \theta$, then $dx = \sec^2 \theta d\theta$. Limits: When $x=0$, $\theta=0$. When $x=1$, $\theta=\frac{\pi}{4}$. Integral becomes: $\int_0^{\pi/4} \frac{\tan^2 \theta}{(1+\tan^2 \theta)^{3/2}} \sec^2 \theta d\theta = \int_0^{\pi/4} \frac{\tan^2 \theta}{(\sec^2 \theta)^{3/2}} \sec^2 \theta d\theta$ $= \int_0^{\pi/4} \frac{\tan^2 \theta}{\sec^3 \theta} \sec^2 \theta d\theta = \int_0^{\pi/4} \frac{\tan^2 \theta}{\sec \theta} d\theta$ $= \int_0^{\pi/4} \frac{\sin^2 \theta / \cos^2 \theta}{1/\cos \theta} d\theta = \int_0^{\pi/4} \frac{\sin^2 \theta}{\cos \theta} d\theta = \int_0^{\pi/4} \frac{1-\cos^2 \theta}{\cos \theta} d\theta$ $= \int_0^{\pi/4} (\sec \theta - \cos \theta) d\theta$ Evaluate: $[\ln|\sec \theta + \tan \theta| - \sin \theta]_0^{\pi/4}$ $= (\ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \sin(\frac{\pi}{4})) - (\ln|\sec(0) + \tan(0)| - \sin(0))$ $= (\ln|\sqrt{2} + 1| - \frac{1}{\sqrt{2}}) - (\ln|1+0| - 0) = \ln(\sqrt{2}+1) - \frac{1}{\sqrt{2}}$ b) Evaluate $\int_0^1 x^3(1-x)^5 dx$ Beta Function: $\int_0^1 x^m (1-x)^n dx = B(m+1, n+1) = \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}$ Here $m=3, n=5$. $B(3+1, 5+1) = B(4, 6) = \frac{\Gamma(4)\Gamma(6)}{\Gamma(10)}$ $= \frac{(4-1)!(6-1)!}{(10-1)!} = \frac{3!5!}{9!} = \frac{(6)(120)}{362880} = \frac{720}{362880} = \frac{1}{504}$ c) Prove that $\int \frac{e^x - e^{-x}}{\sec x} dx = \frac{1}{2} \log\left(\frac{a^2+1}{2}\right)$, $a>0$ This statement appears to be incorrect or incomplete. The integral $\int \frac{e^x - e^{-x}}{\sec x} dx$ does not result in a constant value dependent on 'a'. It seems there might be a typo in the question, possibly a definite integral with specific limits related to 'a', or the integrand/result is misstated. Assuming the question intended to relate hyperbolic functions and $a$: The general form of the integrand is $\int \frac{2\sinh x}{\sec x} dx = \int 2\sinh x \cos x dx$. This is not straightforward to integrate to the given log form. If the problem meant to prove a property of $\sinh x$ or $\cosh x$ related to $a$, the formulation is missing context. Q4) Curve Tracing and Sphere Equation a) Trace the curve $xy^2 = a^2(a-x)$ Symmetry: Symmetric about the x-axis (since $y$ is squared). Origin: Does not pass through the origin. Intercepts: x-intercept: Set $y=0 \implies a^2(a-x)=0 \implies x=a$. Point $(a,0)$. y-intercept: Set $x=0 \implies 0 = a^3$. This is impossible unless $a=0$, but $a$ is usually a constant. No y-intercept if $a \neq 0$. Asymptotes: Vertical: $x=0$ (y-axis) is an asymptote since $y^2 = \frac{a^2(a-x)}{x}$. As $x \to 0^+$, $y^2 \to \infty$. Horizontal: As $x \to \infty$, $y^2 \to -a^2$. No real horizontal asymptote. Region of Existence: For $y^2$ to be real and positive, $\frac{a^2(a-x)}{x} \ge 0$. If $a>0$: $x$ must be positive and $(a-x)$ must be positive or zero. So $0 The curve exists only for $0 Shape: A loop-like curve extending from $x=0$ (asymptote) to $x=a$ (x-intercept), symmetric about the x-axis. It resembles a 'cissoid of Diocles' but is slightly different. b) Trace the curve $r = a(1 + \cos \theta)$ This is a Cardioid. Symmetry: Symmetric about the polar axis ($\theta=0$) because $\cos(-\theta) = \cos \theta$. Origin: Passes through the origin when $r=0$. $a(1+\cos\theta)=0 \implies \cos\theta = -1 \implies \theta = \pi$. Maximum r: $r_{max} = a(1+1) = 2a$ when $\cos\theta=1 \implies \theta=0$. Point $(2a, 0)$. Minimum r: $r_{min} = a(1-1) = 0$ when $\cos\theta=-1 \implies \theta=\pi$. Shape: A heart-shaped curve. Starts at $(2a,0)$, goes through $(a, \pi/2)$, $(0, \pi)$, $(a, 3\pi/2)$, and back to $(2a,0)$. c) Find the equation of sphere passing through the circle $x^2 + y^2 + z^2 = 4, z=0$ meeting the plane $x+2y-z=0$ in a circle of radius 3. Equation of sphere through a circle: The general equation of a sphere passing through the circle $S=0, P=0$ is $S + \lambda P = 0$. Given circle: $x^2 + y^2 + z^2 - 4 = 0$ (S) and $z=0$ (P). Sphere equation: $x^2 + y^2 + z^2 - 4 + \lambda z = 0$. Rearrange: $x^2 + y^2 + (z + \frac{\lambda}{2})^2 - 4 - \frac{\lambda^2}{4} = 0$. Center of sphere $C = (0, 0, -\frac{\lambda}{2})$. Radius of sphere $R = \sqrt{4 + \frac{\lambda^2}{4}}$. Intersection with plane $x+2y-z=0$: The distance $d$ from the center $C(0, 0, -\frac{\lambda}{2})$ to the plane $x+2y-z=0$ is: $d = \frac{|0 + 2(0) - (-\frac{\lambda}{2})|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|\frac{\lambda}{2}|}{\sqrt{1+4+1}} = \frac{|\frac{\lambda}{2}|}{\sqrt{6}} = \frac{|\lambda|}{2\sqrt{6}}$. Radius of intersection circle: Let $r_c$ be the radius of the intersection circle. We are given $r_c=3$. The relationship is $R^2 = d^2 + r_c^2$. $4 + \frac{\lambda^2}{4} = \left(\frac{|\lambda|}{2\sqrt{6}}\right)^2 + 3^2$ $4 + \frac{\lambda^2}{4} = \frac{\lambda^2}{24} + 9$ Multiply by 24: $96 + 6\lambda^2 = \lambda^2 + 216$ $5\lambda^2 = 216 - 96 = 120$ $\lambda^2 = 24 \implies \lambda = \pm \sqrt{24} = \pm 2\sqrt{6}$. Equation of sphere: Substitute $\lambda$ back into $x^2 + y^2 + z^2 - 4 + \lambda z = 0$. $x^2 + y^2 + z^2 \pm 2\sqrt{6}z - 4 = 0$. Q5) Coordinate Geometry a) Trace the curve $r = a \sin 2\theta$ This is a Four-Leaved Rose. Symmetry: About pole: Yes, if $(r, \theta)$ is on curve, $(-r, \theta)$ or $(r, \theta+\pi)$ is too. $a\sin(2(\theta+\pi)) = a\sin(2\theta+2\pi) = a\sin2\theta$. About polar axis: No. About line $\theta=\pi/2$: Yes, $a\sin(2(\pi-\theta)) = a\sin(2\pi-2\theta) = -a\sin2\theta$. So $(r, \pi-\theta)$ corresponds to $(-r, \theta)$. Origin/Pole: Passes through the origin when $r=0$. $a\sin2\theta=0 \implies 2\theta = n\pi \implies \theta = 0, \pi/2, \pi, 3\pi/2$. Maximum r: $r_{max} = a$ when $\sin2\theta = \pm 1$. $2\theta = \pi/2, 3\pi/2, 5\pi/2, 7\pi/2$. $\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$. These are the tips of the four leaves. Shape: Four petals, each of length $a$. The petals are centered along the lines $\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$. b) Find the equation of the cone with vertex at $(1, 2, -3)$, semi-vertical angle $\cos^{-1}\frac{1}{\sqrt{3}}$ and the line $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+1}{-1}$ as axis of the cone. Vertex: $V = (1, 2, -3)$. Axis Direction Ratios: From the line equation, direction vector $\vec{d} = (1, 2, -1)$. Semi-vertical angle: $\alpha = \cos^{-1}\frac{1}{\sqrt{3}}$, so $\cos \alpha = \frac{1}{\sqrt{3}}$. Equation of cone: For a point $P(x,y,z)$ on the cone, the vector $\vec{VP} = (x-1, y-2, z+3)$. The angle $\alpha$ between $\vec{VP}$ and the axis vector $\vec{d}$ must satisfy $\cos \alpha = \frac{\vec{VP} \cdot \vec{d}}{|\vec{VP}||\vec{d}|}$. $\vec{VP} \cdot \vec{d} = (x-1)(1) + (y-2)(2) + (z+3)(-1) = x-1 + 2y-4 - z-3 = x+2y-z-8$. $|\vec{VP}| = \sqrt{(x-1)^2 + (y-2)^2 + (z+3)^2}$. $|\vec{d}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1+4+1} = \sqrt{6}$. $\cos \alpha = \frac{1}{\sqrt{3}}$. So, $\frac{1}{\sqrt{3}} = \frac{x+2y-z-8}{\sqrt{(x-1)^2 + (y-2)^2 + (z+3)^2} \sqrt{6}}$. Square both sides: $\frac{1}{3} = \frac{(x+2y-z-8)^2}{((x-1)^2 + (y-2)^2 + (z+3)^2)(6)}$. $2((x-1)^2 + (y-2)^2 + (z+3)^2) = (x+2y-z-8)^2$. This is the equation of the cone. Expand if required, but this form is generally acceptable. c) Find the equation of right circular cylinder of radius 2 whose axis is given by $\frac{x-1}{2} = \frac{y+2}{3} = \frac{z-4}{6}$ Radius: $R=2$. Axis: Passes through point $A(1, -2, 4)$ and has direction vector $\vec{d} = (2, 3, 6)$. Equation of cylinder: For any point $P(x,y,z)$ on the cylinder, the perpendicular distance from $P$ to the axis must be equal to the radius $R$. Let $\vec{AP} = (x-1, y+2, z-4)$. The formula for perpendicular distance from a point to a line is $R = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$. $|\vec{d}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$. $\vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x-1 & y+2 & z-4 \\ 2 & 3 & 6 \end{vmatrix}$ $= \mathbf{i}(6(y+2) - 3(z-4)) - \mathbf{j}(6(x-1) - 2(z-4)) + \mathbf{k}(3(x-1) - 2(y+2))$ $= \mathbf{i}(6y+12-3z+12) - \mathbf{j}(6x-6-2z+8) + \mathbf{k}(3x-3-2y-4)$ $= (6y-3z+24)\mathbf{i} - (6x-2z+2)\mathbf{j} + (3x-2y-7)\mathbf{k}$. $|\vec{AP} \times \vec{d}|^2 = (6y-3z+24)^2 + (6x-2z+2)^2 + (3x-2y-7)^2$. $R^2 = \frac{|\vec{AP} \times \vec{d}|^2}{|\vec{d}|^2}$. $2^2 = \frac{(6y-3z+24)^2 + (6x-2z+2)^2 + (3x-2y-7)^2}{7^2}$. $4 \times 49 = (6y-3z+24)^2 + (6x-2z+2)^2 + (3x-2y-7)^2$. $196 = (6y-3z+24)^2 + (6x-2z+2)^2 + (3x-2y-7)^2$. Q6) Multiple Integrals a) Change the order of integration $\int_0^2 \int_{2-x}^{2+x} f(x,y) dy dx$ Region of Integration: $0 \le x \le 2$ $2-x \le y \le 2+x$ Boundaries: $x=0$ (y-axis) $x=2$ $y = 2-x \implies x = 2-y$ $y = 2+x \implies x = y-2$ Vertices: When $x=0$, $y=2$. Point $(0,2)$. When $x=2$, $y=0$ (from $y=2-x$) and $y=4$ (from $y=2+x$). Points $(2,0)$ and $(2,4)$. New limits (integrating with respect to x first): The region must be split into two parts: For $0 \le y \le 2$: $x$ goes from $y=2-x \implies x=2-y$ to $x=2$. For $2 \le y \le 4$: $x$ goes from $y=2+x \implies x=y-2$ to $x=2$. Changed integral: $\int_0^2 \int_{2-y}^2 f(x,y) dx dy + \int_2^4 \int_{y-2}^2 f(x,y) dx dy$. b) Find the volume bounded by the cylinder $x^2+y^2=4$ and the plane $y+z=4$ and $z=0$. Region: Cylinder: $x^2+y^2=4$ (a circle of radius 2 in the xy-plane). Planes: $z=0$ (xy-plane) and $y+z=4 \implies z=4-y$. Volume Integral: $V = \iint_R (z_{upper} - z_{lower}) dA$. Here $z_{upper}=4-y$ and $z_{lower}=0$. The region $R$ is the disk $x^2+y^2 \le 4$ in the xy-plane. $V = \iint_{x^2+y^2 \le 4} (4-y) dA$. Polar Coordinates: $x = r\cos\theta, y = r\sin\theta$. $dA = r dr d\theta$. Limits: $0 \le r \le 2$, $0 \le \theta \le 2\pi$. $V = \int_0^{2\pi} \int_0^2 (4 - r\sin\theta) r dr d\theta$ $= \int_0^{2\pi} \int_0^2 (4r - r^2\sin\theta) dr d\theta$ Inner integral: $\int_0^2 (4r - r^2\sin\theta) dr = [2r^2 - \frac{r^3}{3}\sin\theta]_0^2 = (2(2^2) - \frac{2^3}{3}\sin\theta) - 0 = 8 - \frac{8}{3}\sin\theta$. Outer integral: $\int_0^{2\pi} (8 - \frac{8}{3}\sin\theta) d\theta$ $= [8\theta + \frac{8}{3}\cos\theta]_0^{2\pi}$ $= (8(2\pi) + \frac{8}{3}\cos(2\pi)) - (8(0) + \frac{8}{3}\cos(0))$ $= (16\pi + \frac{8}{3}) - (0 + \frac{8}{3}) = 16\pi$. Q7) Area and Center of Gravity a) Find the area of one loop of the curve $r = a \cos 2\theta$ This is a Four-Leaved Rose. For $r = a \cos(n\theta)$, the number of petals is $2n$ if $n$ is even. Here $n=2$, so 4 petals. One loop is formed when $r$ goes from $0$ to $a$ and back to $0$. Set $r=0$: $a \cos 2\theta = 0 \implies 2\theta = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, ...$ So $\theta = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}, ...$ A single loop is traced between $\theta = -\frac{\pi}{4}$ and $\theta = \frac{\pi}{4}$. Area in Polar Coordinates: $A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta$. $A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} (a \cos 2\theta)^2 d\theta = \frac{1}{2} \int_{-\pi/4}^{\pi/4} a^2 \cos^2 2\theta d\theta$. Using $\cos^2 x = \frac{1+\cos 2x}{2}$: $A = \frac{a^2}{2} \int_{-\pi/4}^{\pi/4} \frac{1+\cos 4\theta}{2} d\theta = \frac{a^2}{4} \int_{-\pi/4}^{\pi/4} (1+\cos 4\theta) d\theta$. $= \frac{a^2}{4} [\theta + \frac{\sin 4\theta}{4}]_{-\pi/4}^{\pi/4}$. $= \frac{a^2}{4} \left[ \left(\frac{\pi}{4} + \frac{\sin \pi}{4}\right) - \left(-\frac{\pi}{4} + \frac{\sin (-\pi)}{4}\right) \right]$. $= \frac{a^2}{4} \left[ \left(\frac{\pi}{4} + 0\right) - \left(-\frac{\pi}{4} + 0\right) \right] = \frac{a^2}{4} \left[ \frac{\pi}{4} + \frac{\pi}{4} \right] = \frac{a^2}{4} \left[ \frac{2\pi}{4} \right] = \frac{a^2}{4} \frac{\pi}{2} = \frac{\pi a^2}{8}$. b) Find the centre of gravity of the area bounded by $y^2 = x$ and $x+y=2$. Intersection points: Substitute $x=y^2$ into $x+y=2$: $y^2+y=2 \implies y^2+y-2=0 \implies (y+2)(y-1)=0$. $y=-2 \implies x=(-2)^2=4$. Point $(4,-2)$. $y=1 \implies x=1^2=1$. Point $(1,1)$. Area (A): Integrate with respect to $y$. $x$ goes from $y^2$ to $2-y$. $A = \int_{-2}^1 ((2-y) - y^2) dy$. $= [2y - \frac{y^2}{2} - \frac{y^3}{3}]_{-2}^1$. $= (2(1) - \frac{1^2}{2} - \frac{1^3}{3}) - (2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3})$. $= (2 - \frac{1}{2} - \frac{1}{3}) - (-4 - 2 - (-\frac{8}{3}))$. $= (\frac{12-3-2}{6}) - (-6 + \frac{8}{3}) = \frac{7}{6} - \frac{-18+8}{3} = \frac{7}{6} - (-\frac{10}{3}) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}$. Centroid $(\bar{x}, \bar{y})$: $\bar{x} = \frac{1}{A} \iint_R x dA = \frac{1}{A} \int_{-2}^1 \int_{y^2}^{2-y} x dx dy$. Inner integral: $\int_{y^2}^{2-y} x dx = [\frac{x^2}{2}]_{y^2}^{2-y} = \frac{(2-y)^2}{2} - \frac{(y^2)^2}{2} = \frac{1}{2}(4-4y+y^2-y^4)$. $\bar{x} = \frac{1}{A} \int_{-2}^1 \frac{1}{2}(4-4y+y^2-y^4) dy$. $= \frac{1}{2A} [4y - 2y^2 + \frac{y^3}{3} - \frac{y^5}{5}]_{-2}^1$. $= \frac{1}{2A} \left[ (4-2+\frac{1}{3}-\frac{1}{5}) - (-8-8-\frac{8}{3}+\frac{32}{5}) \right]$. $= \frac{1}{2A} \left[ (\frac{30-15+5-3}{15}) - (\frac{-120-120-40+96}{15}) \right]$. $= \frac{1}{2A} \left[ \frac{17}{15} - \frac{-184}{15} \right] = \frac{1}{2A} \left[ \frac{17+184}{15} \right] = \frac{1}{2A} \frac{201}{15} = \frac{1}{2(\frac{9}{2})} \frac{67}{5} = \frac{1}{9} \frac{67}{5} = \frac{67}{45}$. $\bar{y} = \frac{1}{A} \iint_R y dA = \frac{1}{A} \int_{-2}^1 \int_{y^2}^{2-y} y dx dy$. Inner integral: $\int_{y^2}^{2-y} y dx = [xy]_{y^2}^{2-y} = y(2-y) - y(y^2) = 2y-y^2-y^3$. $\bar{y} = \frac{1}{A} \int_{-2}^1 (2y-y^2-y^3) dy$. $= \frac{1}{A} [y^2 - \frac{y^3}{3} - \frac{y^4}{4}]_{-2}^1$. $= \frac{1}{A} \left[ (1 - \frac{1}{3} - \frac{1}{4}) - (4 - (-\frac{8}{3}) - \frac{16}{4}) \right]$. $= \frac{1}{A} \left[ (\frac{12-4-3}{12}) - (4 + \frac{8}{3} - 4) \right]$. $= \frac{1}{A} \left[ \frac{5}{12} - \frac{8}{3} \right] = \frac{1}{A} \left[ \frac{5-32}{12} \right] = \frac{1}{A} \left[ -\frac{27}{12} \right] = \frac{1}{\frac{9}{2}} \left[ -\frac{9}{4} \right] = \frac{2}{9} \left[ -\frac{9}{4} \right] = -\frac{1}{2}$. Center of Gravity: $(\bar{x}, \bar{y}) = (\frac{67}{45}, -\frac{1}{2})$.