The problem consists of several parts: Question 1 asks to find the derivatives of $f(x) = x^3$ and $g(x) = \sqrt{9-x}$ using the limit definition. Question 2 involves several sub-problems. 2.1 requires proving a derivative formula for $sec^{-1}x$. 2.2 deals with optimizing the area enclosed by a square and an equilateral triangle formed from a 10m wire. 2.3 is a Newton's Law of Cooling problem to determine time of death. 2.4 asks us to differentiate $y = cosh^{-1}(sinh x)$. Question 3 contains five definite integrals to evaluate. (a) $\int \frac{cos(lnx)}{x} dx$ (b) $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx$ (c) $\int_{0}^{\frac{\pi}{2}} \frac{cosx}{1+sin^2x} dx$ (d) $\int_{\frac{1}{2}}^{1} \frac{cos(x^{-2})}{x^3} dx$ (e) $\int_{\sqrt{2}}^{2} \frac{1}{x\sqrt{x^2-1}} dx$

1. Problem Description

The problem consists of several parts:

Question 1 asks to find the derivatives of $f(x) = x^3$ and $g(x) = \sqrt{9-x}$ using the limit definition.

Question 2 involves several sub-problems. 2.1 requires proving a derivative formula for $sec^{-1}x$. 2.2 deals with optimizing the area enclosed by a square and an equilateral triangle formed from a 10m wire. 2.3 is a Newton's Law of Cooling problem to determine time of death. 2.4 asks us to differentiate $y = cosh^{-1}(sinh x)$.

Question 3 contains five definite integrals to evaluate.

(a) $\int \frac{cos(lnx)}{x} dx$

(b) $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx$

(c) $\int_{0}^{\frac{\pi}{2}} \frac{cosx}{1+sin^2x} dx$

(d) $\int_{\frac{1}{2}}^{1} \frac{cos(x^{-2})}{x^3} dx$

(e) $\int_{\sqrt{2}}^{2} \frac{1}{x\sqrt{x^2-1}} dx$

2. Solution Steps

Question 1:

(a) $f(x) = x^3$

$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{(x+h)^3 - x^3}{h} = \lim_{h\to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = \lim_{h\to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h\to 0} (3x^2 + 3xh + h^2) = 3x^2$

(b) $g(x) = \sqrt{9-x}$

$g'(x) = \lim_{h\to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h\to 0} \frac{\sqrt{9-(x+h)} - \sqrt{9-x}}{h}$

Multiply by conjugate: $\lim_{h\to 0} \frac{\sqrt{9-(x+h)} - \sqrt{9-x}}{h} \cdot \frac{\sqrt{9-(x+h)} + \sqrt{9-x}}{\sqrt{9-(x+h)} + \sqrt{9-x}} = \lim_{h\to 0} \frac{9-x-h - (9-x)}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim_{h\to 0} \frac{-h}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim_{h\to 0} \frac{-1}{\sqrt{9-(x+h)} + \sqrt{9-x}} = \frac{-1}{2\sqrt{9-x}}$

Question 2:

2.1

Let $y = sec^{-1}x$, then $x = secy$. Differentiating with respect to $x$:

$1 = secy\cdot tany \cdot \frac{dy}{dx}$.

So, $\frac{dy}{dx} = \frac{1}{secy\cdot tany} = \frac{1}{secy \sqrt{sec^2y - 1}} = \frac{1}{x\sqrt{x^2-1}}$. Since we are given that $|x| > 1$.

2.2

Let $x$ be the length of wire used for the square, and $10-x$ be the length used for the equilateral triangle.

Side of square: $x/4$. Area of square: $(x/4)^2 = x^2/16$.

Side of triangle: $(10-x)/3$. Area of triangle: $\frac{\sqrt{3}}{4} (\frac{10-x}{3})^2 = \frac{\sqrt{3}}{36} (10-x)^2$.

Total Area: $A(x) = \frac{x^2}{16} + \frac{\sqrt{3}}{36}(10-x)^2$

$A'(x) = \frac{2x}{16} + \frac{2\sqrt{3}}{36}(10-x)(-1) = \frac{x}{8} - \frac{\sqrt{3}}{18}(10-x)$

Set $A'(x) = 0$: $\frac{x}{8} = \frac{\sqrt{3}}{18}(10-x) \implies 18x = 8\sqrt{3}(10-x) \implies 18x = 80\sqrt{3} - 8\sqrt{3}x \implies x(18+8\sqrt{3}) = 80\sqrt{3} \implies x = \frac{80\sqrt{3}}{18+8\sqrt{3}} = \frac{40\sqrt{3}}{9+4\sqrt{3}} \approx 4.35 m$.

$A''(x) = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0$. Thus, this is a minimum.

To find the maximum, test endpoints: $x = 0$ or $x=10$.

$A(0) = \frac{\sqrt{3}}{36}(100) = \frac{25\sqrt{3}}{9} \approx 4.81$

$A(10) = \frac{100}{16} = \frac{25}{4} = 6.25$

(a) Maximum when all wire is used for the square (x=10).

(b) Minimum when $x \approx 4.35$.

2.3

Newton's Law of Cooling: $T(t) = T_s + (T_0 - T_s)e^{-kt}$, where $T(t)$ is the temperature at time $t$, $T_s$ is the surrounding temperature, $T_0$ is the initial temperature, and $k$ is a constant.

We have $T_s = 20$. Let $t=0$ be 1:30 PM. Then $T(0) = 32.5$ and $T(1) = 30.3$.

$32.5 = 20 + (37-20)e^{-k(0)} \implies 12.5 = 17$, which is wrong.

The initial temperature of the body should be

3

7. $32.5 = 20 + (37-20)e^{-kt} \implies 12.5 = 17e^{-kt}$

$30.3 = 20 + (37-20)e^{-k(t+1)} \implies 10.3 = 17e^{-k(t+1)}$

$\frac{12.5}{17} = e^{-kt}$ and $\frac{10.3}{17} = e^{-k(t+1)} = e^{-kt}e^{-k}$

$\frac{10.3}{17} = \frac{12.5}{17}e^{-k} \implies \frac{10.3}{12.5} = e^{-k} \implies k = -ln(\frac{10.3}{12.5}) \approx 0.1954$

$\frac{12.5}{17} = e^{-0.1954t} \implies ln(\frac{12.5}{17}) = -0.1954t \implies t = \frac{ln(\frac{12.5}{17})}{-0.1954} \approx 1.525$ hours.

1. 525 hours is about 1 hour and 31.5 minutes. Thus, the murder occurred approximately 1 hour and 31.5 minutes before 1:30 PM, which is about 11:58:30 AM.

2.4

$y = cosh^{-1}(sinh x)$

$\frac{dy}{dx} = \frac{1}{\sqrt{(sinh x)^2 - 1}} \cdot cosh x = \frac{cosh x}{\sqrt{sinh^2 x - 1}}$.

Since $cosh^2 x - sinh^2 x = 1$, we have $cosh^2 x = sinh^2 x + 1$

So we should have

$\frac{dy}{dx} = \frac{1}{\sqrt{sinh^2 x + 1 - 1}} \cdot cosh x = \frac{1}{\sqrt{sinh^2 x}} \cdot cosh x = \frac{cosh x}{cosh x} = \frac{1}{\sqrt{sinh^2x}} cosh(x) = \frac{cosh x}{\sqrt{sinh^2x}}=\frac{cosh(x)}{|sinh(x)|}$.

However, the inverse hyperbolic cosine is only defined for arguments greater than or equal to 1, so $sinh(x)>=1$. Also the hyperbolic cosine is always positive.

$\frac{dy}{dx} = \frac{1}{\sqrt{sinh^2(x) + 1 - 1}} \cdot cosh x = \frac{cosh(x)}{|cosh(x)|} = \frac{cosh(x)}{\sqrt{sinh^2(x) + 1}} = \frac{1}{\sqrt{1 + sinh^2(x) -1}}coshx = \frac{cosh(x)}{\sqrt{sinh^2x}} = \frac{coshx}{|sinhx|}= 1 if sinhx > 0 else -1

Question 3:

(a) $\int \frac{cos(lnx)}{x} dx$. Let $u = lnx$, then $du = \frac{1}{x} dx$.

So, $\int cos(u) du = sin(u) + C = sin(lnx) + C$.

(b) $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx$. Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}}dx \implies 2du = \frac{1}{\sqrt{x}} dx$.

So, $\int e^u 2du = 2e^u + C = 2e^{\sqrt{x}} + C$.

(c) $\int_{0}^{\frac{\pi}{2}} \frac{cosx}{1+sin^2x} dx$. Let $u = sinx$, then $du = cosx dx$. When $x=0$, $u=0$. When $x=\frac{\pi}{2}$, $u=1$.

So, $\int_{0}^{1} \frac{1}{1+u^2} du = [arctan(u)]_{0}^{1} = arctan(1) - arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(d) $\int_{\frac{1}{2}}^{1} \frac{cos(x^{-2})}{x^3} dx$. Let $u = x^{-2}$, then $du = -2x^{-3} dx \implies -\frac{1}{2}du = \frac{1}{x^3} dx$. When $x=\frac{1}{2}$, $u=4$. When $x=1$, $u=1$.

So, $\int_{4}^{1} cos(u) (-\frac{1}{2}) du = -\frac{1}{2}[sin(u)]_{4}^{1} = -\frac{1}{2}(sin(1) - sin(4)) = \frac{1}{2}(sin(4) - sin(1))$.

(e) $\int_{\sqrt{2}}^{2} \frac{1}{x\sqrt{x^2-1}} dx = [sec^{-1}(x)]_{\sqrt{2}}^{2} = sec^{-1}(2) - sec^{-1}(\sqrt{2})$.

$sec^{-1}(2) = \frac{\pi}{3}$ and $sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$.

So, $\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$.

3. Final Answer

Question 1:

(a) $f'(x) = 3x^2$

(b) $g'(x) = \frac{-1}{2\sqrt{9-x}}$

Question 2:

2.1: Shown

2.2: (a) Maximum area is obtained when all the wire is used for the square.

(b) Minimum area is obtained when approximately 4.35 m is used for the square.

2.3: Approximately 11:58:30 AM

2.4: $\frac{cosh(x)}{|sinh(x)|}$.

Question 3:

(a) $sin(lnx) + C$

(b) $2e^{\sqrt{x}} + C$

(c) $\frac{\pi}{4}$

(d) $\frac{1}{2}(sin(4) - sin(1))$

(e) $\frac{\pi}{12}$