Featured 1 week ago

Remember the cubic factoring:

Replace

with the equivalent

and replace

with the equivalent

becomes

Note that as

Since division by zero is undefined, this limit is undefined.

As an aid in verifying this result, here is a graph of

Notice that whether this limit approaches

Featured 1 week ago

I guessed that you meant

The definition of a derivative is:

Given:

Then:

Let's find an equation for h.

Square both sides:

Use the distributive property:

Please observe that

Subtract

Divide both sides by 2:

Substitute this for h into the definition:

Flip the 2 up to the numerartor:

We know how the difference of two squares factors,

Please observe that a common factor cancels:

Here is the equation with the cancelled factors removed:

Now it is ok to let

Featured 1 week ago

Is it:

If so this means that:

We can test if this is **exact** , ie if there exists (level surface) function

With

**exact** and

Next we can look at:

By IBP:

And as:

And as:

Because we know that

So we integrate again:

We know that:

This means that:

**OR:**

Featured 1 week ago

Please see below. Warning: long answer due to explanatory analysis.

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

**Finding the proof**

By definition,

**if and only if**

for every

for all

We have been asked to show that

So we want to make

We want:

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

# = abs(((3-2x) - 6)#

#=abs(-2x-3)#

Recall that we control the size of

I see

# = abs(-2)abs(x+3/2)#

# = 2 abs(x-(-3/2))#

In order to make this less than

**Proving our L is correct -- Writing the proof**

Claim:

Proof:

Given

Now if < delta# then

# = abs(((3-2x) - 6)#

#=abs(-2x-3)#

# = abs(-2)abs(x+3/2)#

# = 2 abs(x-(-3/2))#

# < 2 delta#

# = 2 (epsi/2)#

# = epsilon#

We have shown that for any positive

So, by the definition of limit, we have

**Note**

This is an exampole of a limit in which the strict inequality

Featured 1 week ago

# 4sqrt(3)pia^3 #

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

The red circle has radius

# x^2+y^2=(2a)^2 => x^2+y^2=4a^2#

**Method 1 - Calculate core and subtract from Sphere**

First let us consider the volume of the entire Sphere, which has radius

# :. V_("sphere") = 4/3 pi (2a)^3 #

# " " = 4/3 pi 8a^3 #

# " " = 32/3 pi a^3 #

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about

# V = 2pi \ int_(alpha)^(beta) \ xf(x) \ dx #

Also note that we require twice the volume because we have a portion above and below the

# V_("bore") = 2pi \ int_(0)^(a) \ x(2sqrt(4a^2-x^2)) \ dx #

# " " = 2pi \ int_(0)^(a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let

#u=4a^2-x^2 => (du)/dx = -2x#

When# { (x=0), (x=a) :} => { (u=4a^2), (u=3a^2) :}#

And so:

# V_("bore") = -2pi \ int_(4a^2)^(3a^2) \sqrt(u) \ du #

# " " = 2pi \ int_(3a^2)^(4a^2) \sqrt(u) \ du #

# " " = 2pi \ [ u^(3/2)/(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ ((4a^2)^(3/2) - (3a^2)^(3/2)) #

# " " = 4/3pi \ (8a^3-3sqrt(3)a^3) #

# " " = 32/3pi a^3-4sqrt(3)pia^3 #

And so the total volume is given by:

# V_("total") = V_("sphere") - V_("bore" #

# " " = 32/3 pi a^3 - (32/3pi a^3-4sqrt(3)pia^3)#

# " " = 4sqrt(3)pia^3#

**Method 2 - Calculate volume of bead directly**

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about

# V_("total") = 2pi \ int_(a)^(2a) \ x(2sqrt(4a^2-x^2)) \ dx #

# " " = 2pi \ int_(a)^(2a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let

#u=4a^2-x^2 => (du)/dx = -2x#

When# { (x=a), (x=2a) :} => { (u=3a^2), (u=0) :}#

And so:

# V_("total") = -2pi \ int_(3a^2)^(0) \sqrt(u) \ du #

# " " = 2pi \ int_(0)^(3a^2) \sqrt(u) \ du #

# " " = 2pi \ [ u^(3/2)/(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ ((3a^2)^(3/2) - 0) #

# " " = (4pi)/3 \ (3sqrt(3)a^3 )#

# " " = 4sqrt(3)pia^3 # , as above

Featured 1 week ago

See below.

Calling

we know that the local minima/maxima are given at the stationary points or when

but in our case we have

with

so the stationary points are the solutions of

and the stationary points are

The stationary points qualification is done using

or

or

Regarding the stationary points the value of

so

Attached a plot showing this behavior.

so concluding,

NOTE. The same result for the maximum point would be obtained directly by making

Featured 1 week ago

Let us try a hyperbolic substitution.

Let

Then:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = int color(white)(.)ln(sinh theta + sqrt(sinh^2 theta + 1)) dx/(d theta) d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(sinh theta + cosh theta) cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(e^theta) cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)theta cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = theta sinh theta - cosh theta + C#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = x sinh^(-1) x - sqrt(x^2+1) + C#

If we would prefer not to have an inverse hyperbolic function in the answer, let's find another expression for it:

Let:

#y = sinh^(-1) x#

Then:

#x = sinh y = 1/2(e^y-e^(-y))#

Hence:

#e^y-2x-e^(-y) = 0#

So:

#(e^y)^2-(2x)(e^y)-1 = 0#

So using the quadratic formula:

#e^y = (2x+-sqrt((2x)^2+4))/2#

#color(white)(e^y) = x+-sqrt(x^2+1)#

If

So:

#e^y = x+sqrt(x^2+1)#

and:

#y = ln(x+sqrt(x^2+1))#

That is:

#sinh^(-1) x = ln(x+sqrt(x^2+1))#

So:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C#

Featured yesterday

**Step #1#: Determine the first derivative**

#f'(x) = 6x^2 - 24x + 18#

**Step #2#: Determine the critical numbers**

These will occur when the derivative equals

#0 = 6x^2 - 24x + 18#

#0 = 6(x^2 - 4x + 3)#

#0 = (x - 3)(x - 1)#

#x = 3 or 1#

**Step #3#: Determine the intervals of increase/decrease**

We select test points.

**Test point 1: #x = 0#**

Since this is positive, the function is uniformly increasing on

**Test point 2: #x = 2#**

Since this is negative, the function is decreasing on

I won't select a test point for

**Step #4#: Determine the second derivative **

This is the derivative of the first derivative.

#f''(x) = 12x - 24#

**Step #5#: Determine the points of inflection**

These will occur when

#0 = 12x- 24#

#0 = 12(x - 2)#

#x = 2#

**Step #6#: Determine the intervals of concavity**

Once again, we select test points.

**Test point #1#: #x = 1# **

#f''(1) = 12(1) - 24 = -12#

This means that

This also means that

**Step #7#: Determine the x/y- intercept**

The y-intercept is

#f(0) = 2(0)^3 - 12(0)^2 + 18x - 1 = -1#

If you can't connect the graph, you could always make a table of values. In the end, you should get a graph similar to the following. graph{2x^3- 12x^2 + 18x - 1 [-10, 10, -5, 5]}

Hopefully this helps!

Featured 1 week ago

For the **Proof,** refer to the **Explanation.**

Here is a **Second Method** to prove the **Result** for

Now,

Knowing that,

Rest of the Soln. is the same as in the **First Method.**

**Enjoy Maths.!**

Featured 6 days ago

A height of

Let

The formula for surface area of a rectangular prism is generally:

#S.A = 2lh + 2lw + 2hw#

In this case, it becomes

#600 = 2lh+ 2l^2 + 2lh#

#600 = 4lh + 2l^2#

We want to solve for one of the variables now.

#600 - 2l^2 = 4lh#

#(600 -2l^2)/(4l) =h#

#(300 - l^2)/(2l) = h#

The volume of a rectangular prism is given by:

#V =l xx w xx h#

Which in our case becomes:

#V = l^2 xx h#

Now:

#V = l^2(300 - l^2)/(2l)#

#V = 1/2l(300 - l^2)#

#V = 1/2(300l - l^3)#

#V = 150l - 1/2l^3#

Now find the derivative of this with respect to volume.

#V' = 150 - 3/2l^2#

Find the critical values. These will occur when the derivative equals

#0 = 150 - 3/2l^2#

#3/2l^2 = 150#

#l^2 = 100#

#l = +- 10#

A negative side length doesn't make sense, so we don't accept it as a solution. We now verify graphically to insure this is a maximum.

And it is!

We realize the y-coordinate of the maximum is

#V = l^2 * h#

#1000 = (10)^2 * h#

#1000 = 100h#

#h = 10#

Hopefully this helps!