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## How do I find the time, 𝑇(𝑥), that is required to travel from A to C via P?

Steve
Featured 1 month ago

$D \left(x\right) = \sqrt{9 + {x}^{2}} + \left(6 - x\right)$

$T \left(x\right) = \frac{\sqrt{9 + {x}^{2}}}{4} + \frac{6 - x}{5}$

From this we get the minimum time as $1.65$ (hours) which corresponds to a distance of $7$ (km).

#### Explanation:

A) The distance $D \left(x\right)$ from A to C via P as a function of $x$.

By Pythagoras;

$\setminus \setminus \setminus \setminus \setminus A {P}^{2} = A {B}^{2} + B {P}^{2}$
$\therefore A {P}^{2} = {3}^{2} + {x}^{2}$
$\therefore A {P}^{2} = 9 + {x}^{2}$
$\therefore \setminus \setminus A P = \sqrt{9 + {x}^{2}}$ (must be the +ve root)

Then;

$\setminus \setminus \setminus \setminus \setminus D \left(x\right) = A P + P C$
$\therefore D \left(x\right) = \sqrt{9 + {x}^{2}} + \left(6 - x\right)$

B) Find the time, $T \left(x\right)$, that is required to travel from A to C via P.

Using $\text{speed" = "distance"/"time}$:

Along AP the speed is $4$ km/hour, provided $x > 0$ (otherwise we are going along AB at $5$ km/hour) and so:

$\text{ } 4 = \frac{\sqrt{9 + {x}^{2}}}{t} _ 1$
$\therefore {t}_{1} = \frac{\sqrt{9 + {x}^{2}}}{4}$

Along PC (or AB) the speed is $5$ km/hour, and so:

$\text{ } 5 = \frac{6 - x}{t} _ 2$
$\therefore {t}_{2} = \frac{6 - x}{5}$

And so, the total time is given by:

$\setminus \setminus \setminus \setminus \setminus T \left(x\right) = {t}_{1} + {t}_{2}$
$\therefore T \left(x\right) = \frac{\sqrt{9 + {x}^{2}}}{4} + \frac{6 - x}{5}$ for ($x > 0$)

C) Stevie's Quest

In order for Stevie to complete the quest we need to find a critical point of T(x):

Differentiating wrt $x$ we get:

$\text{ } T ' \left(x\right) = \frac{1}{4} \cdot \frac{1}{2} {\left(9 + {x}^{2}\right)}^{- \frac{1}{2}} \cdot 2 x + \frac{1}{5} \left(- 1\right)$
$\therefore T ' \left(x\right) = \frac{x}{4 \sqrt{\left(9 + {x}^{2}\right)}} - \frac{1}{5}$

At a critical point, $T ' \left(x\right) = 0$

$\implies \frac{x}{4 \sqrt{9 + {x}^{2}}} - \frac{1}{5} = 0$
$\therefore 5 x - 4 \sqrt{9 + {x}^{2}} = 0$
$\therefore 5 x = 4 \sqrt{9 + {x}^{2}}$
$\therefore 25 {x}^{2} = 16 \left(9 + {x}^{2}\right)$
$\therefore 25 {x}^{2} = 144 + 16 {x}^{2}$
$\therefore 9 {x}^{2} = 144$
$\therefore {x}^{2} = \frac{144}{9}$
$\therefore {x}^{2} = 16$
$\therefore x = 4$ (must be the +ve root)

When $x = 4$ we have:

$\therefore D \left(4\right) = \sqrt{9 + 16} + \left(6 - 4\right) = 7$
$\therefore T \left(4\right) = \frac{\sqrt{9 + 16}}{4} + \frac{6 - 4}{5} = \frac{33}{20} = 1.65$

We can confirm visually that this corresponds to a minimum by looking at the graph:

graph{sqrt(9 + x^2)/4 + (6-x)/5 [-15, 15, -1, 10]}

## What is the Particular solution of: dy/dx-3ycotx=sin2x If y=2, when x=π/2?

Ratnaker Mehta
Featured 1 month ago

$\text{The General Solution : } y + 2 {\sin}^{2} x = c {\sin}^{3} x .$

$\text{The Particular Solution : } y + 2 {\sin}^{2} x = 4 {\sin}^{3} x .$

#### Explanation:

This is a Linear Differential Equation of First Order , usually

taken as, $\frac{\mathrm{dy}}{\mathrm{dx}} + y P \left(x\right) = Q \left(x\right)$.

To find its General Soln. , we need to multiply it by the Integrating

Factor, i.e., IF , given by, IF $= {e}^{\int P \left(x\right) \mathrm{dx}}$.

In our Example, $P \left(x\right) = - 3 \cot x , s o , \int P \left(x\right) \mathrm{dx} = \int \left\{- 3 \cot x\right\} \mathrm{dx}$

$= - 3 \ln \sin x = \ln {\sin}^{- 3} x \Rightarrow {e}^{P \left(x\right) \mathrm{dx}} = {e}^{\ln {\sin}^{- 3} x} = \frac{1}{\sin} ^ 3 x$.

Multiplying the Diff. Eqn., by $\frac{1}{\sin} ^ 3 x$, we get,

${\sin}^{- 3} x \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y \cot \frac{x}{\sin} ^ 3 x = \frac{\sin 2 x}{\sin} ^ 3 x$

$\therefore {\sin}^{- 3} x \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y \cos x {\sin}^{- 4} x = \frac{2 \sin x \cos x}{\sin} ^ 3 x = \frac{2 \cos x}{\sin} ^ 2 x$

$\therefore {\sin}^{- 3} x \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left({\sin}^{- 3} x\right) = 2 \cot x \csc x$

$\therefore \frac{d}{\mathrm{dx}} \left\{y {\sin}^{- 3} x\right\} = 2 \cot x \csc x$

Integrating, $y {\sin}^{- 3} x = 2 \int \cot x \csc x + c = - 2 \csc x + c$

$\therefore y + 2 {\sin}^{2} x = c {\sin}^{3} x ,$ is the reqd. Gen. Soln.

To find the Particular Soln. , we utilise the Initial Condition :

$y = 2 , \text{ when } x = \frac{\pi}{2}$.

Sub.ing in the Gen. Soln., we get, $2 + 2 = c = 4$

Hene, the P.S. $: y + 2 {\sin}^{2} x = 4 {\sin}^{3} x$

Enjoy Maths.!

## How do you use the ratio test to test the convergence of the series #∑ (n!)^2 / ( (kn)!)# from n=1 to infinity?

Andrea S.
Featured 3 weeks ago

The series:

#sum_(n=1)^oo ((n!)^2)/((kn)!)#

is convergent for $k \ge 2$

#### Explanation:

The ratio test states that a necessary condition for a series ${\sum}_{n = 1}^{\infty} {a}_{n}$ to converge is that:

$L = {\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid \le 1$

if $L < 1$ the condition is also sufficient.

In our case:

#abs (a_(n+1)/a_n) = ((((n+1)!)^2)/((k(n+1))!))/(((n!)^2)/((kn)!)) = (((n+1)!)^2)/((n!)^2) ((kn)!)/ ((kn+k)!) = (n+1)^2/((kn+k) (kn+k-1)...(kn+1))#

Now we have that:

1. For $k = 1$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} {\left(n + 1\right)}^{2} / \left(n + 1\right) = {\lim}_{n \to \infty} \left(n + 1\right) = + \infty$

and the series is divergent.

1. For $k = 2$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} {\left(n + 1\right)}^{2} / \left(\left(n + 2\right) \left(n + 1\right)\right) = 1$

and the test is inconclusive, so we have to look at the series in more detail:

#sum_(n=1)^oo ((n!)^2)/((2n)!) = (n(n-1)(n-2)...2*1)/ (2n(2n-1)...(n+1)#

We can note that the numerator has $n$ factors from $1$ to $n$ and the denominator has $n$ factors from $n + 1$ to $2 n$, so ordering them appropriately we have:

$\frac{n \left(n - 1\right) \left(n - 2\right) \ldots 2}{2 n \left(2 n - 1\right) \ldots \left(n + 1\right)} = {\prod}_{q = 1}^{n} \frac{q}{n + q}$

Now as $q \le n$,

$\frac{q}{n + q} \le \frac{q}{q + q} = \frac{1}{2}$

So we have:

$\frac{n \left(n - 1\right) \left(n - 2\right) \ldots 2}{2 n \left(2 n - 1\right) \ldots \left(n + 1\right)} \le {\left(\frac{1}{2}\right)}^{n}$

And as:

${\sum}_{n = 1}^{\infty} {\left(\frac{1}{2}\right)}^{n} = 1$

is convergent, then also our series is convergent by direct comparison.

1. For $k \ge 3$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} {\left(n + 1\right)}^{2} / \left(\left(n + 1\right) \left(n + 2\right) \ldots \left(n + k\right)\right) = 0$

and the series is convergent.

## How do you find the positive values of p for which #Sigma lnn/n^p# from #[2,oo)# converges?

Andrea S.
Featured 6 days ago

The series:

${\sum}_{n = 2}^{\infty} \ln \frac{n}{n} ^ p$

is convergent for $p > 1$.

#### Explanation:

A necessary condition for a series to converge is that its general term is infinitesimal, that is:

(1) ${\lim}_{n \to \infty} \ln \frac{n}{n} ^ p = 0$

Consider the function $f \left(x\right) = \ln \frac{x}{x} ^ p$ and the limit:

(2) ${\lim}_{x \to \infty} \ln \frac{x}{x} ^ p$

Clearly if such limit exists it must coincide with (1).

Now, for $p = 0$ the limit (2) is $\infty$, while for $p \ne 0$ it is in the indeterminate form $\frac{\infty}{\infty}$ so we can solve it using l'Hospital's rule:

${\lim}_{x \to \infty} \ln \frac{x}{x} ^ p = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} {x}^{p}} = {\lim}_{x \to \infty} \left(\frac{1}{x}\right) \left(\frac{1}{p {x}^{p - 1}}\right) = {\lim}_{x \to \infty} \frac{1}{p {x}^{p}} = \left\{\begin{matrix}0 \text{ for " p > 0 \\ oo " for } p < 0\end{matrix}\right.$

So we know that for $p \le 0$ the series is not convergent.

To have a sufficient condition we can apply the integral test, using as test function: $f \left(x\right) = \ln \frac{x}{x} ^ p$.

For $p > 0$ such function is positive, decreasing and, as we have just seen, infinitesimal, so it satisfies the hypotheses of the integral test theorem, and the series is proven to be convergent if the improper integral:

${\int}_{2}^{\infty} \ln \frac{x}{x} ^ p$

also converges.

Solving the indefinite integral by parts:

$\int \ln \frac{x}{x} ^ p \mathrm{dx} = \int \ln x d \left({x}^{1 - p} / \left(1 - p\right)\right) = \ln x {x}^{1 - p} / \left(1 - p\right) - \int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx} = \frac{1}{1 - p} \ln \frac{x}{x} ^ \left(p - 1\right) - \int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx}$

$\int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx} = \frac{1}{p - 1} \int {x}^{- p} \mathrm{dx} = \frac{1}{1 - p} ^ 2 {x}^{1 - p} + C = \frac{1}{1 - p} ^ 2 \frac{1}{x} ^ \left(p - 1\right) + C$

So that:

${\int}_{2}^{\infty} \ln \frac{x}{x} ^ p = {\left[\frac{1}{1 - p} \ln \frac{x}{x} ^ \left(p - 1\right) - \frac{1}{1 - p} ^ 2 \frac{1}{x} ^ \left(p - 1\right)\right]}_{2}^{\infty}$

which is convergent for $p - 1 > 0 \implies p > 1$

In conclusion the series is convergent for $p > 1$

## Does #a_n=x^n/(xn!) # converge for any x?

Cesareo R.
Featured 2 weeks ago

converges for $x \in {\mathbb{R}}^{+}$

#### Explanation:

Using the Stirling asymptotic approximation

#n! approx (n/e)^n# we have

#x^n/((x cdot n)!) approx x^n/((x cdot n)/e)^(x cdot n)=(x/x^x(e/n)^x)^n#

The sequence converges if $\frac{x}{x} ^ x {\left(\frac{e}{n}\right)}^{x} < 1$ so

$\frac{e}{n} < {x}^{\frac{x - 1}{x}}$ Now given a $n$ such that $\frac{e}{n} \le \epsilon$

then the convergence is attained for ${x}^{\frac{x - 1}{x}} > \epsilon$

The function $f \left(x\right) = {x}^{\frac{x - 1}{x}}$ for $x \in {\mathbb{R}}^{+}$ has a minimum at $x = 1$ and $f \left(1\right) = 1$. Concluding the sequence is convergent for $x \in {\mathbb{R}}^{+}$

Attached a plot of $f \left(x\right)$ for $x \in {\mathbb{R}}^{+}$

## How do you use the integral test to determine if #Sigma n^ke^-n# from #[1,oo)# where k is an integer is convergent or divergent?

Andrea S.
Featured 6 days ago

The series

${\sum}_{= 1}^{\infty} {n}^{k} {e}^{- n}$

is convergent for any integer $k$

#### Explanation:

Choose as test function:

$f \left(x\right) = {x}^{k} {e}^{- x}$

This function is:

(i) non negative, as $f \left(x\right) > 0$ for $x > 0$

(ii) strictly decreasing in $\left(k , + \infty\right)$ as:

$f ' \left(x\right) = k {x}^{k - 1} {e}^{- x} - {x}^{k} {e}^{- x} = {x}^{k - 1} {e}^{- x} \left(k - x\right) < 0$ for $x > k$

(iii) infinitesimal, as:

${\lim}_{x \to \infty} {x}^{k} {e}^{- x} = 0$

(iv) $f \left(n\right) = {n}^{k} {e}^{- n}$

so the convergence of the series is equivalent to the convergence of the integral:

${\int}_{1}^{\infty} {x}^{k} {e}^{- x} \mathrm{dx}$

We have to find a general formula for this integral. Start from $k = 1$, integrating by parts:

${I}_{1} = {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} x d \left({e}^{- x}\right) = {\left[- x {e}^{- x}\right]}_{1}^{\infty} + {\int}_{1}^{\infty} {e}^{- x} \mathrm{dx} = \frac{1}{e} - {\left[{e}^{- x}\right]}_{1}^{\infty} = \frac{2}{e}$

For $k = 2$

${I}_{2} = {\int}_{1}^{\infty} {x}^{2} {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} {x}^{2} d \left({e}^{- x}\right) = {\left[- {x}^{2} {e}^{- x}\right]}_{1}^{\infty} + 2 {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = \frac{1}{e} + 2 {I}_{1} = \frac{5}{e}$

and in general:

${I}_{k} = {\int}_{1}^{\infty} {x}^{k} {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} {x}^{k} d \left({e}^{- x}\right) = {\left[- {x}^{k} {e}^{- x}\right]}_{1}^{\infty} + k {\int}_{1}^{\infty} {x}^{k - 1} {e}^{- x} \mathrm{dx} = \frac{1}{e} + k {I}_{k - 1}$

#I_k= 1/e + kI_(k-1) = 1/e +k(1/e+(k-1))I_(k-2) = ... = (1+k+k(k-1)+...+k!)/e = 1/esum_(j=1)^k (k!)/(j!)#

So we can assert that:

#int_1^oo x^ke^(-x)dx = 1/esum_(j=1)^k (k!)/(j!)#

is convergent for every $k$ and this proves that also:

${\sum}_{= 1}^{\infty} {n}^{k} {e}^{- n}$

is convergent.

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