Question

2024: section three
a. find the derivative of the following. you do not need to simplify your answer.
i. f(x)=e^{2x} sl = 1
ii. f(x)=x^{2}e^{x^{2}} sl = 2
b. consider the curve given by the function f(x)=x^{3}-3x^{2}+2x. determine the equation of the tangent line to the curve at the point where x = 1. sl = 3
c. calculate the stationary point of f(x)=\frac{x^{2}}{1 - x}, and determine the nature of f at such point. sl = 3
d. describe the condition where \lim_{x\to a}f(x) exists. sl = 2
e. the first derivative of the parametric equations x = \theta-\sin\theta and y = 1 - \cos\theta. is \frac{dy}{dx}=\frac{\sin\theta}{1 - \cos\theta}. find the second derivative \frac{d^{2}y}{dx^{2}} in terms of \theta. sl = 2
f. use implicit differentiation to find \frac{dy}{dx} if x^{2}+y^{2}=\sin(xy). sl = 3
g. sketch a graph of f(x) on the interval -3 < x < 3 that satisfies the following conditions.
i. differentiable at 0 < x < 3 and -3 < x < - 1
ii. discontinuous at x = 0
iii. increasing on -3 < x < -1
iv. decreasing on -1 < x < 3
v. stationary point at x = 2
vi. continuous but not differentiable at x=-1 sl = 3
h. an ice - sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. the volume of the sphere is decreasing at a constant rate of 2\pi cubic meters per hour. calculate at what rate in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 5 meters. (note: for a sphere of radius r, the surface area is 4\pi r^{2} and the volume is \frac{4}{3}\pi r^{3}) sl = 4

Explanation:

Step1: Find derivative of $f(x)=e^{2x}$

Use chain - rule. Let $u = 2x$, then $y = e^{u}$. $\frac{dy}{du}=e^{u}$ and $\frac{du}{dx}=2$. So $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2e^{2x}$.

Step2: Find derivative of $f(x)=x^{2}e^{x^{2}}$

Use product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x^{2}$ and $v = e^{x^{2}}$. $u^\prime=2x$, and for $v = e^{x^{2}}$, let $t=x^{2}$, then $\frac{dv}{dt}=e^{t}$ and $\frac{dt}{dx}=2x$, so $\frac{dv}{dx}=2xe^{x^{2}}$. Then $f^\prime(x)=2xe^{x^{2}}+x^{2}\cdot2xe^{x^{2}}=2xe^{x^{2}}(1 + x^{2})$.

Step3: Find equation of tangent line for $y=x^{3}-3x^{2}+2x$ at $x = 1$

First, find the derivative $y^\prime=3x^{2}-6x + 2$. When $x = 1$, $y^\prime(1)=3-6 + 2=-1$. When $x = 1$, $y=1-3 + 2=0$. Using the point - slope form $y - y_{1}=m(x - x_{1})$, the equation of the tangent line is $y-0=-1(x - 1)$, i.e., $y=-x + 1$.

Step4: Find stationary point of $f(x)=\frac{x^{2}}{1 - x}$

Use quotient - rule $(\frac{u}{v})^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}$, where $u = x^{2}$, $u^\prime = 2x$, $v = 1 - x$, $v^\prime=-1$. So $f^\prime(x)=\frac{2x(1 - x)-x^{2}(-1)}{(1 - x)^{2}}=\frac{2x-2x^{2}+x^{2}}{(1 - x)^{2}}=\frac{2x - x^{2}}{(1 - x)^{2}}$. Set $f^\prime(x)=0$, then $2x - x^{2}=x(2 - x)=0$, so $x = 0$ or $x = 2$.

Step5: Determine the nature of stationary points

Find the second - derivative of $f(x)$ using the quotient - rule on $f^\prime(x)=\frac{2x - x^{2}}{(1 - x)^{2}}$. After simplification, when $x = 0$, $f^{\prime\prime}(0)=2>0$, so it's a local minimum. When $x = 2$, $f^{\prime\prime}(2)=-2<0$, so it's a local maximum.

Step6: Describe the condition for $\lim_{x

ightarrow a}f(x)$ to exist
The left - hand limit $\lim_{x
ightarrow a^{-}}f(x)$ and the right - hand limit $\lim_{x
ightarrow a^{+}}f(x)$ must exist and be equal, i.e., $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)$.

Step7: Find the second derivative of parametric equations

Given $\frac{dy}{dx}=\frac{\sin\theta}{1 - \cos\theta}$, use the formula $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{d\theta}(\frac{dy}{dx})}{\frac{dx}{d\theta}}$. $\frac{dx}{d\theta}=1-\cos\theta$, and $\frac{d}{d\theta}(\frac{\sin\theta}{1 - \cos\theta})=\frac{\cos\theta(1 - \cos\theta)-\sin\theta\sin\theta}{(1 - \cos\theta)^{2}}=\frac{\cos\theta - 1}{(1 - \cos\theta)^{2}}=-\frac{1}{1 - \cos\theta}$. So $\frac{d^{2}y}{dx^{2}}=-\frac{1}{(1 - \cos\theta)^{2}}$.

Step8: Use implicit differentiation for $x^{2}+y^{2}=\sin(xy)$

Differentiate both sides with respect to $x$: $2x + 2y\frac{dy}{dx}=\cos(xy)\cdot(y+x\frac{dy}{dx})$. Then $2x + 2y\frac{dy}{dx}=y\cos(xy)+x\cos(xy)\frac{dy}{dx}$. Rearrange to get $\frac{dy}{dx}=\frac{y\cos(xy)-2x}{2y - x\cos(xy)}$.

Step9: Sketch the graph for $f(x)$

Based on the given conditions:
- Differentiable on $0\lt x\lt3$ and $-3\lt x\lt - 1$ means the graph has a smooth curve in these intervals.
- Discontinuous at $x = 0$ implies a break at $x = 0$.
- Increasing on $-3\lt x\lt - 1$ means the slope is positive in this interval.
- Decreasing on $-1\lt x\lt3$ means the slope is negative in this interval.
- Stationary point at $x = 2$ means the slope is 0 at $x = 2$.
- Continuous but not differentiable at $x=-1$ means a sharp turn at $x=-1$.

Step10: Solve related - rates problem for the sphere

The volume of a sphere $V=\frac{4}{3}\pi r^{3}$, $\frac{dV}{dt}=-2\pi$. Differentiate $V$ with respect to $t$: $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$. When $r = 5$, $-2\pi=4\pi\times5^{2}\frac{dr}{dt}$, so $\frac{dr}{dt}=-\frac{1}{50}$. The surface area $S = 4\pi r^{2}$, differentiate with respect to $t$: $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$. Substitute $r = 5$ and $\frac{dr}{dt}=-\frac{1}{50}$, we get $\frac{dS}{dt}=8\pi\times5\times(-\frac{1}{50})=-\frac{4}{5}\pi$.

Answer:

a. i. $2e^{2x}$; ii. $2xe^{x^{2}}(1 + x^{2})$
b. $y=-x + 1$
c. Stationary points at $x = 0$ (local minimum) and $x = 2$ (local maximum)
d. $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)$
e. $-\frac{1}{(1 - \cos\theta)^{2}}$
f. $\frac{y\cos(xy)-2x}{2y - x\cos(xy)}$
g. Sketch as per the described properties
h. $-\frac{4}{5}\pi$ square meters per hour