Question

2023: section three

1. differentiate the following functions. you do not need to simplify your answer.

i. (f(x)=\tan2x) sl = 1
ii. (y = lnsqrt{2x + 1}) sl = 2
iii. (g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}) sl = 3

1. find (lim_{x

ightarrowinfty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}) sl = 2

1. the graph of a piece - wise function (f(x)) is shown below.

i. find (lim_{x
ightarrow2}f(x)) sl = 1
ii. sketch the graph of the derivative of (f(x)). sl = 2

1. use implicit differentiation to find (\frac{dy}{dx}) if (x^{2}+y^{2}-2y = 1) sl = 2
2. the curve traced by a point on a circle as it rolls on a straight line has parametric equations (x=\theta-sin\theta,y = 1-cos\theta). find (\frac{d^{2}y}{dx^{2}}) in terms of (\theta). sl = 3
3. a 10 m ladder is leaning against a vertical wall. the angle, (\theta), formed by the ladder and the ground changes if the top of the ladder slides down the wall at a constant rate of 2 m/s, what rate of the angle (\theta) changing when the top of the ladder is 6 m above the ground sl = 3
4. a cone is inscribed in a sphere of radius 3 cm, centered at o. the height of the cone is (h) and the radius of the base is (r). find the height of the cone so that the cone has a maximum volume. sl = 4

Explanation:

1. Differentiate \(f(x)=\tan(2x)\)

Use the chain - rule. If \(y = \tan(u)\) and \(u = 2x\), the derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^{2}(u)\), and the derivative of \(u = 2x\) with respect to \(x\) is \(2\).
\[f^\prime(x)=2\sec^{2}(2x)\]

2. Differentiate \(y = \ln\sqrt{2x + 1}\)

First, rewrite \(y=\ln(2x + 1)^{\frac{1}{2}}=\frac{1}{2}\ln(2x + 1)\). Then, using the chain - rule, if \(y=\frac{1}{2}\ln(u)\) and \(u = 2x+1\), the derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\), and the derivative of \(u = 2x + 1\) with respect to \(x\) is \(2\).
\[y^\prime=\frac{1}{2}\times\frac{2}{2x + 1}=\frac{1}{2x+1}\]

3. Differentiate \(g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}\)

Use the quotient - rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = e^{2x}\), \(u^\prime=2e^{2x}\), \(v=(x^{2}+1)^{10}\), and \(v^\prime = 10(x^{2}+1)^{9}\times2x\).
\[g^\prime(x)=\frac{2e^{2x}(x^{2}+1)^{10}-e^{2x}\times20x(x^{2}+1)^{9}}{(x^{2}+1)^{20}}\]

4. Find \(\lim_{x

ightarrow\infty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}\)
Divide both the numerator and denominator by \(x^{3}\):
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}&=\lim_{x ightarrow\infty}\frac{\frac{1}{x^{3}}+\frac{2}{x}-1}{3-\frac{2}{x^{2}}+\frac{1}{x^{3}}}\\ &=\frac{0 + 0-1}{3-0 + 0}=-\frac{1}{3} \end{align*} \]

5. For the piece - wise function \(f(x)\) (graph given):

i. Find \(\lim_{x

ightarrow2}f(x)\)
Examine the left - hand limit and right - hand limit as \(x ightarrow2\) from the graph. If the left - hand limit \(\lim_{x ightarrow2^{-}}f(x)\) and the right - hand limit \(\lim_{x ightarrow2^{+}}f(x)\) are equal, then \(\lim_{x ightarrow2}f(x)\) exists.

ii. Sketch the graph of \(f^\prime(x)\)

Analyze the slope of \(f(x)\) at different intervals from the graph of \(f(x)\). Where \(f(x)\) is increasing, \(f^\prime(x)>0\); where \(f(x)\) is decreasing, \(f^\prime(x)<0\); and at the points of horizontal tangency, \(f^\prime(x) = 0\).

6. Use implicit differentiation for \(x^{2}+y^{2}-2y = 1\)

Differentiate both sides with respect to \(x\):
\[2x + 2y\frac{dy}{dx}-2\frac{dy}{dx}=0\]
\[2y\frac{dy}{dx}-2\frac{dy}{dx}=-2x\]
\[\frac{dy}{dx}=\frac{-2x}{2y - 2}=\frac{x}{1 - y}\]

7. For parametric equations \(x=\theta-\sin\theta\), \(y = 1-\cos\theta\)

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

8. For the ladder problem

Let \(y\) be the height of the top of the ladder on the wall. We know \(y = 6\) m, the length of the ladder \(L = 10\) m. By the Pythagorean theorem, \(x^{2}+y^{2}=100\), and \(\sin\theta=\frac{y}{10}\).
Differentiate \(\sin\theta=\frac{y}{10}\) with respect to time \(t\): \(\cos\theta\frac{d\theta}{dt}=\frac{1}{10}\frac{dy}{dt}\).
When \(y = 6\), \(x=\sqrt{100 - 36}=8\), so \(\cos\theta=\frac{8}{10}\) and \(\frac{dy}{dt}=-2\) m/s.
\(\frac{8}{10}\frac{d\theta}{dt}=\frac{1}{10}\times(-2)\)
\(\frac{d\theta}{dt}=-\frac{1}{4}\) rad/s

9. For the cone inscribed in a sphere

The radius of the sphere \(R = 3\) cm. Using the Pythagorean relationship \(r^{2}=R^{2}-(h - R)^{2}=9-(h - 3)^{2}=6h - h^{2}\).
The volume of the cone \(V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(6h - h^{2})h=\frac{1}{3}\pi(6h^{2}-h^{3})\).
Differentiate \(V\) with respect to \(h\): \(V^\prime=\frac{1}{3}\pi(12h - 3h^{2})\).
Set \(V^\prime = 0\): \(12h-3h^{2}=0\), \(3h(4 - h)=0\). So \(h = 4\) cm.

Answer:

1. i. \(2\sec^{2}(2x)\)

ii. \(\frac{1}{2x + 1}\)
iii. \(\frac{2e^{2x}(x^{2}+1)^{10}-20xe^{2x}(x^{2}+1)^{9}}{(x^{2}+1)^{20}}\)

1. \(-\frac{1}{3}\)
2. i. Value of limit from graph analysis

ii. Sketch of derivative graph from original graph analysis

1. \(\frac{x}{1 - y}\)
2. \(-\frac{1}{(1 - \cos\theta)^{2}}\)
3. \(-\frac{1}{4}\) rad/s
4. \(h = 4\) cm