Calculus I — Limits & Derivatives

Question 1 [10]

(a) By similar triangles the water surface radius $r$ satisfies $\frac{r}{h}=\frac{6}{12}=\frac12$, so $r=\frac{h}{2}$. [1] Volume of cone of water: $V=\frac13\pi r^2 h = \frac13\pi\left(\frac{h}{2}\right)^2 h = \frac13\pi\frac{h^2}{4}h=\frac{\pi h^3}{12}$. [2]

(b) Net rate of volume change: $\frac{dV}{dt}= 3 - \frac14 h$. At $h=4$: $\frac{dV}{dt}=3-1=2$ m³/min. [2] Differentiate $V=\frac{\pi h^3}{12}$: $\frac{dV}{dt}=\frac{\pi \cdot 3h^2}{12}\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$. [2] At $h=4$: $\frac{dV}{dt}=\frac{\pi(16)}{4}\frac{dh}{dt}=4\pi\frac{dh}{dt}$. [1] So $2 = 4\pi\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{2\pi}\approx 0.159$ m/min. [2]

Why: related rates — differentiate the geometric relation w.r.t. $t$, substitute the net (inflow − leak) rate.

Question 2 [10]

(a) $\lim_{x\to0^-}\frac{\sin 3x}{x} = \lim \frac{\sin 3x}{3x}\cdot 3 = 1\cdot 3 = 3$ (using $\lim_{u\to0}\frac{\sin u}{u}=1$). [3]

(b) Continuity at $0$: left limit $=3$; value $= a(0)+b = b$. So $b=3$. [1]

(c) Rationalise: multiply by $\frac{\sqrt x+\sqrt2}{\sqrt x+\sqrt2}$: $\frac{(x^2-4)(\sqrt x+\sqrt2)}{x-2}.$ Factor $x^2-4=(x-2)(x+2)$: $=(x+2)(\sqrt x+\sqrt2).$ [3] As $x\to2^+$: $(2+2)(\sqrt2+\sqrt2)= 4\cdot 2\sqrt2 = 8\sqrt2$. [1] Left value at $x=2$: $f(2)=2a+b$. Right limit $=\frac{8\sqrt2}{c}$. Continuity: $\frac{8\sqrt2}{c}=2a+b \Rightarrow c=\frac{8\sqrt2}{2a+b}$. [2]

Why: removable-type factor cancellation; matching one-sided limits to the middle piece's value.

Question 3 [10]

(a) We want: $\forall \varepsilon>0,\exists\delta>0$ s.t. $0<|x-3|<\delta\Rightarrow |x^2-2x-3|<\varepsilon$. [1] Factor: $x^2-2x-3=(x-3)(x+1)$, so $|x^2-2x-3|=|x-3||x+1|$. [1] Restrict $\delta\le 1$: then $|x-3|<1\Rightarrow 2<x<4\Rightarrow 3<x+1<5$, so $|x+1|<5$. [2] Thus $|x-3||x+1|<5|x-3|<5\delta$. Choose $\delta=\min\{1,\varepsilon/5\}$. [1] Then $|x^2-2x-3|<5\cdot\frac{\varepsilon}{5}=\varepsilon$. ∎ [1]

(b) Since $-1\le\cos(1/x^3)\le 1$ for $x\ne0$, multiply by $x^2\ge0$: $-x^2 \le x^2\cos(1/x^3)\le x^2.$ [2] As $x\to0$, both $-x^2\to0$ and $x^2\to0$. By Squeeze Theorem the limit is $0$. [2]

Question 4 [10]

(a) Differentiate: $3x^2 + 3y^2 y' = 6y + 6x y'$. [2] $3y^2y' - 6xy' = 6y - 3x^2 \Rightarrow y'(3y^2-6x)=6y-3x^2$. $y'=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}.$ [2]

(b) Horizontal tangent: numerator $=0$ (with denominator $\ne0$): $2y = x^2 \Rightarrow y=\frac{x^2}{2}$. [2] Substitute into $x^3+y^3=6xy$: $x^3 + \frac{x^6}{8} = 6x\cdot\frac{x^2}{2}=3x^3.$ $\frac{x^6}{8} = 2x^3 \Rightarrow x^6 = 16 x^3 \Rightarrow x^3(x^3-16)=0.$ [2] First quadrant: $x^3=16\Rightarrow x=16^{1/3}=2\cdot2^{1/3}=2^{4/3}$. Then $y=\frac{x^2}{2}=\frac{2^{8/3}}{2}=2^{5/3}$. Point: $\left(2^{4/3},\,2^{5/3}\right)\approx(2.52,\,3.17)$. (Denominator $y^2-2x=2^{10/3}-2^{7/3}\ne0$, valid.) [2]

Question 5 [10]

(a) $g'(x)=e^{-x/2} + x\cdot(-\tfrac12)e^{-x/2} = e^{-x/2}\left(1-\tfrac{x}{2}\right)$. [2] Since $e^{-x/2}>0$, sign determined by $1-\frac x2$: $g'>0$ for $x<2$, $g'<0$ for $x>2$. So $g'$ changes $+\to-$ at $x=2$: local maximum. [1] $g(2)=2e^{-1}=\frac{2}{e}\approx 0.736$. [1]

(b) $g''(x)=\frac{d}{dx}\left[e^{-x/2}(1-\tfrac x2)\right] = -\tfrac12 e^{-x/2}(1-\tfrac x2) + e^{-x/2}(-\tfrac12)$ $= e^{-x/2}\left(-\tfrac12+\tfrac x4-\tfrac12\right)=e^{-x/2}\left(\tfrac x4 - 1\right)$. [1] $g''=0$ at $x=4$; $g''<0$ for $x<4$ (concave down), $g''>0$ for $x>4$ (concave up). [1] Inflection point: $\left(4,\,4e^{-2}\right)\approx(4,0.541)$. [1]

(c) $\lim_{x\to\infty} x e^{-x/2}=\lim_{x\to\infty}\frac{x}{e^{x/2}}$ — form $\frac{\infty}{\infty}$, apply L'Hôpital: $=\lim\frac{1}{\tfrac12 e^{x/2}}=0$. [2] End behaviour: as $x\to\infty$, $g\to 0^+$; the $x$-axis ($y=0$) is a horizontal asymptote; graph rises to max at $x=2$ then decays to $0$. [1]

[
  {"claim":"Q1: dh/dt = 1/(2pi) at h=4","code":"h=symbols('h'); pi=pi; V=pi*h**3/12; dVdh=diff(V,h); dhdt=Rational(2,1)/dVdh.subs(h,4); result= simplify(dhdt-1/(2*pi))==0"},
  {"claim":"Q2c: limit of (x^2-4)/(sqrt x - sqrt2) at x->2 is 8*sqrt2","code":"x=symbols('x'); L=limit((x**2-4)/(sqrt(x)-sqrt(2)),x,2); result= simplify(L-8*sqrt(2))==0"},
  {"claim":"Q4b: first-quadrant horizontal-tangent point (2**(4/3), 2**(5/3)) lies on folium","code":"xx=2**Rational(4,3); yy=2**Rational(5,3); result= simplify(xx**3+yy**3-6*xx*yy)==0"},
  {"claim":"Q5: g' zero at x=2 (local max) and g'' zero at x=4 (inflection)","code":"x=symbols('x'); g=x*exp(-x/2); g1=diff(g,x); g2=diff(g,x,2); result= (simplify(g1.subs(x,2))==0) and (simplify(g2.subs(x,4))==0)"}
]