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Featured 2 months ago

# dy/dx = 2/(3x^2-3)#

# y = ln((x-1)/(x-1))^(1/3) #

# :. y = 1/3ln((x-1)/(x+1)) #

# :. 3y = ln((x-1)/(x+1)) #

# :. e^(3y) = ((x-1)/(x+1)) # ..... [1]

Differentiating the LHS implicity and the RHS using the quotient rule gives:

# :. 3e^(3y)dy/dx = ( (x+1)(1) - (x-1)(1) )/(x+1)^2#

# :. 3e^(3y)dy/dx = ( x+1-x+1 )/(x+1)^2#

# :. 3e^(3y)dy/dx = 2/(x+1)^2#

# :. 3((x-1)/(x+1))dy/dx = 2/(x+1)^2# (using [1])

# :. dy/dx = 2/3*1/(x+1)^2*(x+1)/(x-1)#

# :. dy/dx = 2/3*1/(x+1)*1/(x-1)#

# :. dy/dx = 2/3*1/(x^2-1)#

Hence,

# dy/dx = 2/(3x^2-3)#

Featured 1 month ago

Let

Let

Then

Now we simply plug into the formula with a starting point.

Second iteration:

And so on, starting with

Featured 2 weeks 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 week ago

It doesn't. But it does satisfy the conclusion. See below.

There are two hypotheses for MVT.

The function must be continuous on

The function must be differentiable on

The first is true (satisfied) because

This function fails to be differentiable at

This function does not satisfy the hypotheses of the Mean Value Theorem on this interval.

**Bonus material**

This function DOES satisfy the conclusion of the MVT on this interval. We cannot use the Mean Value Theorem to conclude that there is a

We can, however solve

The algebra is be tedious to type, but the graph makes this plausible.

It shows

graph{(y-x^(1/3)) (y-4^{1/3)-((4^(1/3)+5^(1/3))/9)(x-4))= 0 [-6.06, 5.04, -2.864, 2.685]}

Featured 6 days ago

Our goal here is to get rid of the

#=int 2/((5sectheta)^4sqrt((5sectheta)^2 - 25)) * 5secthetatanthetad theta#

Apply the identity

#=int 2/(625sec^4thetasqrt(25tan^2theta))* 5secthetatantheta d theta#

#=int 2/(625sec^4(5tantheta)) * 5secthetatantheta d theta#

#=int 2/(625sec^3theta) d theta#

Rewrite using

#=int 2/625cos^3theta d theta#

#=int 2/625 cos^2theta(costheta) d theta#

Rearrange using the pythagorean identity

#=int 2/625 (1 - sin^2theta)costheta d theta#

Let

#=int 2/625 (1 - u^2)costheta * (du)/costheta#

#=int 2/625( 1 - u^2) du#

Integrate using

#=2/625(u - 1/3u^3) + C#

#=2/625sintheta - 2/1875sin^3theta + C#

Draw a triangle to represent

#=2/625sqrt(y^2 - 25)/y - 2/1875(sqrt(y^2 - 25)/y)^3 + C#

#=(2sqrt(y^2 - 25))/(625y) - (2(y^2-25)^(3/2))/(1875y^3) + C#

Hopefully this helps!

Featured 2 days ago

Because the order of the numerator is greater than the denominator some reduction must be done before expanding the partial fractions. Please see the explanation.

Given:

Begin reducing by adding 0 to the numerator in the form of

Break into two fractions:

Factor

The first term becomes

Add 0 to the numerator of the second term in the form of

Break the second term into two fractions:

Remove a factor of 3 from the first numerator and combine like terms in the last fraction:

The second term becomes 3:

Partial Fraction Expansion of the last term:

Make B disappear by letting x = -2

Make A disappear by Letting x = 1:

Check

This checks

Returning to the main problem:

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