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Featured 1 month ago

From this we get the minimum time as

**A) The distance #D(x)# from A to C via P as a function of #x#.**

By Pythagoras;

# \ \ \ \ \ AP^2 = AB^2 + BP^2 #

# :. AP^2 = 3^2 + x^2 #

# :. AP^2 = 9 + x^2 #

# :. \ \ AP = sqrt(9 + x^2) # (must be the +ve root)

Then;

# \ \ \ \ \ D(x) = AP + PC #

# :. D(x) = sqrt(9 + x^2) + (6-x) #

**B) Find the time, #T(x)#, that is required to travel from A to C via P.**

Using

Along AP the speed is

# " "4 = sqrt(9 + x^2)/t_1 #

# :. t_1 = sqrt(9 + x^2)/4 #

Along PC (or AB) the speed is

# " " 5 = (6-x)/t_2 #

# :. t_2 = (6-x)/5 #

And so, the total time is given by:

# \ \ \ \ \ T(x) = t_1 + t_2 #

# :. T(x) = sqrt(9 + x^2)/4 + (6-x)/5 # for (#x gt 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

# " "T'(x) = 1/4*1/2(9+x^2)^(-1/2)*2x+1/5(-1) #

# :. T'(x) = x/(4sqrt((9+x^2))) - 1/5 #

At a critical point,

# => x/(4sqrt(9+x^2)) - 1/5 = 0 #

# :. 5x - 4sqrt(9+x^2) = 0 #

# :. 5x = 4sqrt(9+x^2) #

# :. 25x^2 = 16(9+x^2) #

# :. 25x^2 = 144 + 16x^2 #

# :. 9x^2 = 144 #

# :. x^2 = 144/9 #

# :. x^2 = 16 #

# :. x = 4 # (must be the +ve root)

When

# :. D(4) = sqrt(9 + 16) + (6-4) = 7#

# :. T(4) = sqrt(9 + 16)/4 + (6-4)/5 = 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]}

Featured 1 month ago

This is a **Linear Differential Equation** of **First Order** , usually

taken as,

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

Factor, i.e., **IF** , given by, **IF**

In our Example,

Multiplying the Diff. Eqn., by

Integrating,

**Gen. Soln.**

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

Sub.ing in the Gen. Soln., we get,

Hene, the **P.S.**

Enjoy Maths.!

Featured 3 weeks ago

The series:

is convergent for

The ratio test states that a necessary condition for a series

if

In our case:

Now we have that:

- For
#k= 1#

and the series is divergent.

- For
#k= 2#

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

We can note that the numerator has

Now as

So we have:

And as:

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

- For
#k >= 3#

and the series is convergent.

Featured 6 days ago

The series:

is convergent for

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

(1)

#lim_(n->oo) lnn/n^p= 0#

Consider the function

(2)

#lim_(x->oo) lnx/x^p#

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

Now, for

So we know that for

To have a sufficient condition we can apply the integral test, using as test function:

For

also converges.

Solving the indefinite integral by parts:

So that:

which is convergent for

In conclusion the series is convergent for

Featured 2 weeks ago

converges for

Using the Stirling asymptotic approximation

The sequence converges if

then the convergence is attained for

The function

Attached a plot of

Featured 6 days ago

The series

is convergent for any integer

Choose as test function:

This function is:

(i) non negative, as

#f(x) > 0# for#x > 0# (ii) strictly decreasing in

#(k,+oo)# as:

#f'(x) = kx^(k-1)e^(-x) -x^ke^(-x) = x^(k-1)e^(-x) (k-x) < 0# for#x > k# (iii) infinitesimal, as:

#lim_(x->oo) x^ke^(-x) = 0# (iv)

#f(n) = n^ke^(-n)#

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

We have to find a general formula for this integral. Start from

For

and in general:

So we can assert that:

is convergent for every

is convergent.

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