Curvature of Curves

Max.: $x\approx -2.19$, $y\approx 24.19$; min.:, $x\approx 1.52$, $y\approx -1.38$.

$$\begin{array}{c}y=x^3+x^2-10x+8\\ \frac{dy}{dx}=3x^2+2x-10\\ \frac{dy}{dx}=0\Rightarrow x=\frac{-2\pm \sqrt{4+120}}{6}=\frac{-2\pm 2\sqrt{31}}{6}\\ \text{ Therefore }\frac{dy}{dx}=0\text{ if }x=-\frac{1+\sqrt{31}}{3}\approx -2.189\text{ or }x=\frac{\sqrt{31}-1}{3}\approx 1.522\end{array}$$

To distinguish between a maximum and minimum, we find the second derivative:

$$\frac{d^{2}y}{d{x}^2}=6x+2$$

When $x\approx -2.19$ Therefore, the curve is concave downward close to $x\approx -2.19$ and $x\approx -2.19$ corresponds to a maximum $y\approx 24.19.$

When $x\approx 1.52$ \begin{align} & y \approx-1.38 \\ & \frac{d^{2} y}{d x^{2}}=6 \times 1.52+2>0 \end{align} Therefore, the curve is concave upward near $x\approx 1.52$ and $x\approx 1.52$ corresponds to a minimum $y\approx -1.38$.

This curve is shown below:

${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}-2cx$; ${\displaystyle \frac{d^{2}y}{d{x}^2}}=-2c$; $x={\displaystyle \frac{b}{2ac}}$ (a maximum).

\begin{align} y & =\frac{b}{a} x-c x^{2} \\ \frac{d y}{d x} & =\frac{b}{a}-2 c x=0 \Rightarrow x=\frac{b}{2 a c} \\ \frac{d^{2} y}{d x^{2}} & =-2 c \end{align}

When $x=\frac{b}{2ac}$

$$y=\frac{b}{a}\cdot \frac{b}{2ac}-c\frac{b^2}{4a^2{c}^2}=\frac{b^2}{4a^{2}c}$$ This is a maximum value since .

(a) One maximum and two minima.
(b) One maximum. ($x=0$; other points unreal.)

\begin{align} & y=1-\frac{x^{2}}{2}+\frac{x^{4}}{24} \\ & \frac{d y}{d x}=-x+\frac{1}{6} x^{3}=x\left(\frac{x^{2}}{6}-1\right) \\ & \frac{d y}{d x}=0 \quad \Leftrightarrow \quad x=0 \quad \text { or } \quad x= \pm \sqrt{6} \\ & \frac{d^{2} y}{d x^{2}}=-1+\frac{1}{2} x^{2} \end{align}

When $x=0$, Therefore, $x=0$ makes $y$ a maximum value. When $x=0$, $y=1$.

When $x=\sqrt{6}$ or $x=-\sqrt{6}$ Therefore, $x=\sqrt{6}\approx 2.449$ or $x=-\sqrt{6}\approx -2.449$ corresponds to a minimum $y=-{\displaystyle \frac{1}{2}}$.

The graph of $y=1-\frac{x^2}{2}+\frac{x^4}{24}$ is shown below:

Now let’s consider the second function $$y=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}.$$ \begin{align} \frac{d y}{d x} & =-x+\frac{x^{3}}{6}-\frac{x^{5}}{120}=\frac{1}{120} x\left(-120+20 x^{2}-x^{4}\right) \\ \frac{d y}{d x}=0 & \Leftrightarrow x=0 \text { or } x^{4}-20 x^{2}+120=0 \end{align}

The expression $x^4-20x^2+120$ is quadratic in terms of $t=x^2$

$$t^2-20t+120=0$$

Since the discriminant of this equation, ${20}^2-4\times 120=-80$, is negative, the above equation has no roots.

Therefore,

$$\frac{dy}{dx}=0\iff x=0$$

The value of $y$ for $x=0$ is 1 . Since

$$\frac{d^{2}y}{d{x}^2}=-1+\frac{x^2}{2}-\frac{x^4}{24}$$

is negative when $x=0$, $y=1$ is a maximum.

The graph of $y=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$ is shown below:

Min.: $x\approx 1.71$, $y\approx 6.13$.

$$y=2x+1+\frac{5}{x^2}$$ We can rewrite it as $$y=2x+1+5x^{-2}$$ Then $$\frac{dy}{dx}=2-10x^{-3}=2-\frac{10}{x^3}$$ The second derivative is: $$\frac{d^{2}y}{d{x}^2}=30x^{-4}=\frac{30}{x^4}$$

To find where $y$ is a maximum or minimum we set ${\displaystyle \frac{dy}{dx}}$ equal to zero: $$\frac{dy}{dx}=0\iff x=\sqrt[3]{5}$$

Since , $x=\sqrt[3]{5}$ corresponds to a minimum $$y=2\sqrt[3]{5}+1+\frac{5}{5^{\frac{2}{3}}}\approx 6.13.$$

Max: $x=-.5$, $y=4$.

$$y=\frac{3}{x^2+x+1}$$

Using the Quotient Rule

\begin{align} & \frac{d y}{d x}=-\frac{3(2 x+1)}{\left(x^{2}+x+1\right)^{2}} \\ & \frac{d y}{d x}=0 \quad \Leftrightarrow \quad x=-\frac{1}{2} \end{align}

To determine if $y$ has a maximum or a minimum when $x=-\frac{1}{2}$, we can use the Second Derivative Test:

$$\frac{d^{2}y}{d{x}^2}=-3\frac{2{(x^2+x+1)}^2-2(2x+1{)}^2(x^2+x+1)}{{(x^2+x+1)}^4}.$$

When $x=-\frac{1}{2}$

$$\frac{d^{2}y}{d{x}^2}=-3\frac{2(+)-2\times 0(\dots )}{(+)}=(-)$$ Therefore, the curve is concave downward and $x=-\frac{1}{2}$ corresponds to a maximum $y=4$.

Alternatively, we can use the First Derivative Test.

When , , the curve ascends

When , , the curve descends.

Thus $y=4$ is a maximum that occurs when $x=-\frac{1}{2}$.

The graph of $y={\displaystyle \frac{3}{x^2+x+1}}$ is shown below.

Max.: $x\approx 1.414$, $y\approx 1.768$.
Min.: $x\approx -1.414$, $y\approx 1.768$.

$$y=\frac{5x}{2+x^2}$$

Using the Quotient Rule:

$$\frac{dy}{dx}=\frac{5(2+x^2)-5x(2x)}{{(2+x^2)}^2}$$

or

$$\frac{dy}{dx}=\frac{10-5x^2}{{(2+x^2)}^2}$$

\begin{align} & \frac{d y}{d x}=0 \Leftrightarrow 10-5 x^2=0 \\ & \frac{d y}{d x}=0 \Leftrightarrow \quad x= \pm \sqrt{2} \end{align}

Using the Second Derivative Test: $$\frac{d^{2}y}{d{x}^2}=\frac{-10x{(2+x^2)}^2-2(2x)(2+x^2)(10-5x^2)}{{(2+x^2)}^4}$$

When $x=\sqrt{2}$ $$\frac{d^{2}y}{d{r}^2}=\frac{-10(+)(+)-2(+)(+)(0)}{(+)}=(-)$$ It follows from the Second Derivative Test that $x=\sqrt{2}$ corresponds to a maximum $y={\displaystyle \frac{5\sqrt{2}}{2+2}}\approx 1.768$.

When $x=-\sqrt{2}$ $$\frac{d^{2}y}{d{x}^2}=\frac{-10(-)(+)-2(+)(+)(0)}{(+)}=(+)$$ This means that $x=-\sqrt{2}$ corresponds to a minimum $y=-{\displaystyle \frac{5\sqrt{2}}{4}}\approx -1.768$.

The graph of $y={\displaystyle \frac{5x}{2+x^2}}$ is shown below.

Max.: $x\approx -3.565$, $y\approx 2.12$.
Min.: $x\approx +3.565$, $y\approx 7.88$.

\begin{align} & y=\frac{3 x}{x^{2}-3}+\frac{x}{2}+5 \\ \frac{d y}{d x} & =\frac{3\left(x^{2}-3\right)-6 x^{2}}{\left(x^{2}-3\right)^{2}}+\frac{1}{2} \\ & =\frac{-9-3 x^{2}}{\left(x^{2}-3\right)^{2}}+\frac{1}{2} \\ & =\frac{2\left(-9-3 x^{2}\right)+\left(x^{2}-3\right)^{2}}{2\left(x^{2}-3\right)^{2}} \\ & =\frac{-18-6 x^{2}+x^{4}-6 x^{2}+9}{2\left(x^{2}-3\right)^{2}} \\ & =\frac{x^{4}-12 x^{2}-9}{2\left(x^{2}-3\right)^{2}} \end{align} $$\frac{dy}{dx}=0\iff x^4-12x^2-9=0.$$ The equation $x^4-12x^2-9=0$ is quadratic in terms of $x^2$. Thus $$x^2=\frac{12\pm \sqrt{144+36}}{2}=\frac{12\pm \sqrt{180}}{2}$$ The negative is unacceptable because $x^2\ge 0$. Therefore

$$\begin{array}{c}{x}^2=\frac{12+\sqrt{180}}{2}=3(\sqrt{5}+2)\\ \frac{dy}{dx}=0\iff x=\sqrt{3(\sqrt{5}+2)}\approx 3.565\text{ or }x=-\sqrt{3(\sqrt{5}+2)}\approx -3.565\end{array}$$

Using the Second Derivative Test $$\frac{d^{2}y}{d{x}^2}=\frac{2(3x^2-24x){(x^2-3)}^2-8x(x^2-3)(x^4-12x^2-9)}{4{(x^2-3)}^4}$$

When $x=\sqrt{3(\sqrt{5}+2)}\approx 3.565$

\begin{align} \frac{d^{2} y}{d x^{2}} & =\frac{3\left(3 \times 3.565^{2}-24 \times 3.565\right)(+)-8(+)(-)(0)}{(+)} \\ & =\frac{3(-)(+)-0}{(+)} \\ & =(-) \end{align} Therefore, $x\approx 3.565$ corresponds to a maximum $y\approx 7.884$.

When $x\approx -3.565$

$$\frac{d^{2}y}{d{x}^2}=\frac{3(+)(+)-0}{(+)}=(+)$$

Therefore, $x\approx -3.565$ corresponds to a minimum $y\approx 2.116$.

The graph of $y={\displaystyle \frac{3x}{x^2-3}}+{\displaystyle \frac{x}{2}}+5$ is shown below.

$0.4N$, $0.6N$.

Let

$x=$ part one

$z=$ part two

We know $$x+z=N$$ and we want to minimize $$3x^2+2z^2$$

Since $z=N-x$, we want to minimize

$$3x^2+2(N-x{)}^2$$

Let $$y=3x^2+2(N-x{)}^2$$ then \begin{align} \frac{d y}{d x} & =6 x+2 \times 2 \times(-1)(N-x) \\ & =10 x-4 N \end{align} $$\frac{dy}{dx}=0\iff x=0.4N$$

Does $x=0.4$ correspond to a minimum $y$ or a maximum $y$? To answer this, we can use the Second Derivative Test

It follows from the Second Derivative Test that $x=0.4N$ corresponds to a minimum value.

When $x=0.4N$, $z=0.6N$ and

\begin{align} & y=3(0.4 N)^{2}+2(0.6 N)^{2} \\ & y=1.2 N^{2} \end{align}

$x=\sqrt{{\displaystyle \frac{a}{c}}}$.

$$u=\frac{x}{a+bx+c{x}^2}$$

Using Quotient Rule:

\begin{align} \frac{d u}{d x} & =\frac{\left(a+b x+c x^{2}\right)-x(b+2 c x)}{\left(a+b x+c x^{2}\right)} \\ & =\frac{a-c x^{2}}{\left(a+b x+c x^{2}\right)^{2}} \end{align} $$\frac{du}{dx}=0\iff x=\pm \sqrt{\frac{c}{a}}$$ $$\frac{d^{2}u}{d{x}^2}=\frac{-2cx(a+bx+cx{)}^2+(a-c{x}^2)(b+2cx)}{{(a+bx+c{x}^2)}^4}$$

When $x=\sqrt{{\displaystyle \frac{c}{a}}}$, $$\frac{d^{2}u}{d{x}^2}=\frac{-2c\sqrt{\frac{c}{a}}(+)+0}{(+)}=(-)$$ [Note that $a-c{x}^2$ is zero when $x=\sqrt{\frac{c}{a}}$ ]

Therefore, $x=\sqrt{{\displaystyle \frac{c}{a}}}$ makes $u$ a maximum.

$x$ cannot be negative (What is the meaning of negative output of an electric generator?), but even if were acceptable, when $x=-\sqrt{\frac{c}{a}}$, we have

$$\frac{d^{2}u}{d{x}^2}=\frac{-2c(-\sqrt{\frac{c}{a}})(+)+0}{(+)}=(+),$$ Therefore $u$ is a minimum when $x=-\sqrt{{\displaystyle \frac{c}{a}}}$.

Speed $\sqrt[3]{650}\approx 8.66$ nautical miles per hour. Time taken $115.44$ hours.
Minimum cost $\mathrm{\$}29714.7$.

$$\frac{\text{no. of tons}}{\text{hr}}=y=0.3+0.001v^3$$ Since the cost of other expenses is equivalent to $1\frac{\text{ton of coal}}{\text{hr}}$, if the voyage takes $t$ hours, then the total cost of the voyage is $$\text{cost }=a(y\cdot t+t)=a(y+1)t$$ where $a$ is the cost of coal per ton.

If the speed of the steamer is $v$, since the voyage is $1000$ nautical miles, the time of the voyage is $$t=\frac{1000}{v}$$ Therefore, we can write that the cost of the voyage is \begin{align} \text{cost } &=a\left(1.3+0.001v^3\right)\frac{1000}{v}\\ &=a\left(\frac{1300}{v}+v^2\right). \end{align}

To minimize the cost, we differentiate cost with respect to velocity and set the result equal to zero:

$$\frac{d\text{ cost}}{dv}=a(-\frac{1300}{v^2}+2v)=0$$

$$\Rightarrow 2v=\frac{1300}{v^2}$$ $$\Rightarrow v=\sqrt[3]{650}\approx 8.662$$

8.662 nautical miles per hour is the speed that will make the total cost a minimum.

If the steamer moves by the speed $\sqrt[3]{650}$, the voyage takes ${\displaystyle \frac{1000}{\sqrt[3]{650}}}\approx 115.442$ hours.

To find the minimum cost, we have to calculate $\text{cost}=a({\displaystyle \frac{1300}{v}}+v^2)$ for $a=132$ and $v=\sqrt[3]{650}\approx 8.66$: $$\text{cost}\approx 132(\frac{1300}{8.662}+{8.662}^2)\approx 29714.7.$$

Max. and min. for $x=7.5$, $y\approx \pm 5.413$.

First consider $$y=\frac{x}{6}\sqrt{x(10-x)}$$ Then $$\frac{dy}{dx}=\frac{1}{6}\sqrt{x(10-x)}+\frac{x}{6}\frac{d\left\{\sqrt{x(10-x)}\right\}}{dx}$$

To differentiate $\sqrt{x(10-x)}$, let $u=x(10-x)$ and \begin{align} \frac{d \left(\sqrt{u}\right)}{d x} & =\frac{d \left(\sqrt{u}\right)}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{2 \sqrt{u}} \cdot(10-2 x) \\ & =\frac{1}{\sqrt{x(10-x)}} \cdot(5-x) \end{align} Therefore,

\begin{align} \frac{d y}{d x}&=\frac{1}{6} \sqrt{x(10-x)}+\frac{x}{6} \frac{5-x}{\sqrt{x(10-x)}} \\ &=\frac{x(10-x)+x(5-x)}{6 \sqrt{x(10-x)}} \\ &=\frac{15 x-2 x^{2}}{6 \sqrt{x(10-x)}} \end{align} \begin{align} \frac{d y}{d x}=0 \quad &\Leftrightarrow\quad x(15-2 x)=0 \\ &\Leftrightarrow \quad x=0 \text { or } x=7.5 \end{align}

To distinguish between a maximum or a minimum, we need the sign of the second derivative for $x=0$ and for $x=7.5$.

$$\frac{d^{2}y}{d{x}^2}=\frac{1}{6}\frac{(15-4x)\sqrt{x(10-x)}-\frac{d(\sqrt{x(10-x)})}{dx}\cdot (15x-2x^2)}{(\sqrt{x(10-x)}{)}^2}$$

Note that, to find the sign of $\frac{d^{2}y}{d{x}^2}$, we do not need to write the expression for $\frac{d(\sqrt{x(10-x)})}{dx}$ because the result will be multiplied by $(15x-2x^2)$ which is zero for both $x=0$ and $x=7.5$.

When $x=0$ Therefore, $x=0$ corresponds to a minimum $y=0$.

When $x=7.5$ Hence $x=7.5$ corresponds to a maximum $$y=\frac{7.5}{6}\sqrt{7.5\times (10-7.5)}\approx 5.413$$

If we are careless, we might say that $x=0$ correspom to a minimum $y=0$, but we notice that $y={\displaystyle \frac{x}{6}}\sqrt{x(10-x)}$ is if (it is not defined for ). However, this curve has another branch $y=-{\displaystyle \frac{x}{6}}\sqrt{x(10-x)}$ which is when . Therefore, the curve has neither a minimum nor a maximum if $x=0$.

If we consider $y=-{\displaystyle \frac{x}{6}}\sqrt{x(10-x)}$, its derivative is also zero when $x=0$ or $x=7.5$. The second derivative has the opposite sign of the second derivative of the other branch $y={\displaystyle \frac{x}{6}}\sqrt{x(10-x)}$. Therefore, when $x=7.5$. Hence $y=-{\displaystyle \frac{x}{6}}\sqrt{x(10-x)}$ has a minimum value of $y\approx -5.413$ when $x=7.5$.

The curve $y=\pm {\displaystyle \frac{x}{6}}\sqrt{x(10-x)}$ is shown below.

Min.: $x=\frac{1}{2}$, $y=0.25$; max.: $x=-\frac{1}{3}$, $y\approx 1.407$.

\begin{align} y & =4 x^{3}-x^{2}-2 x+1 \\ \frac{d y}{d x} & =12 x^{2}-2 x-2=2\left(6 x^{2}-x-1\right) \\ \frac{d y}{d x}=0 &\quad \Leftrightarrow \quad x=\frac{1 \pm \sqrt{1+24}}{12}=\frac{1 \pm 5}{12} \\ \frac{d y}{d x} =0 & \quad \Leftrightarrow \quad x=\frac{1}{2} \quad \text { or } \quad x=-\frac{1}{3} \\ \frac{d^{2} y}{d x^{2}} & =2(12 x-1) \end{align}

When $x={\displaystyle \frac{1}{2}}$, , the curve is concave upward and $y$ has a minimum value of $4\times {\left({\displaystyle \frac{1}{2}}\right)}^3-{\left({\displaystyle \frac{1}{2}}\right)}^2-2\left({\displaystyle \frac{1}{2}}\right)+1={\displaystyle \frac{1}{4}}$ when $x={\displaystyle \frac{1}{2}}$.

When $x=-{\displaystyle \frac{1}{3}}$, , the curve is concave downward and $y$ has a maximum value of ${\displaystyle \frac{38}{27}}\approx 1.407$ when $x=-{\displaystyle \frac{1}{3}}$.

The graph of $y=4x^3-x^2-2x+1$ is shown below.