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Featured 2 months ago

# â„’^-1 {s/(s^4+4a^4)} \ \ \ \ \ = (sin(at)sinh(at))/(2a^2) #

# â„’^-1 {s/(s^4+s^2+1)} = 2/sqrt(3)sin(sqrt(3)/2)sinh(1/2t) #

**Part (A)**

First we can complete the square:

# s/(s^4+4a^4) = s/( (s^2+2a^2)^2 - 4a^2s^2) #

# " " = s/( (s^2+2a^2)^2 - (2as)^2) #

We now have the difference of two squares, so:

# s/(s^4+4a^4) = s/( (s^2+2a^2 + 2as)(s^2+2a^2 - 2as) ) #

Now let us form the partial fraction decomposition for the expression, which will be of the form:

# s/(s^4+4a^4) = A/(s^2+2a^2 + 2as) + B/(s^2+2a^2 - 2as) #

# " " = (A(s^2+2a^2 - 2as) + B(s^2+2a^2 + 2as))/ ((s^2+2a^2 + 2as)(s^2+2a^2 - 2as)) #

# :. s = A(s^2+2a^2 - 2as) + B(s^2+2a^2 + 2as) #

Equating Coefficients we get:

# Coeff(s) \ \ => 1 = -2aA+2aB \ \ \ \ \ ..... [1]#

# Coeff(s^2) => 0 = A+B \ \ \ \ \ ..... [2]#

Now we solve these simultaneous equations:

# Eq[1] + (2a)Eq[2] => 1=4aB #

# :. A=-1/(4a), B= 1/(4a) #

Se we can write the original expression as:

# s/(s^4+4a^4) = 1/(4a) { 1/(s^2+2a^2 - 2as) - 1/(s^2+2a^2 + 2as) } #

# " " = 1/(4a) { 1/((s-a)^2-a^2+2a^2) - 1/((s+a)^2-a^2+2a^2) } #

# " " = 1/(4a) { 1/((s-a)^2+a^2) - 1/((s+a)^2+a^2) } #

And we can now take inverse Laplace Transforms using standard results to get:

# â„’^-1 {s/(s^4+4a^4)} = 1/(4a){e^(at)sin(at)/a - e^(-at)sin(at)/a} #

# " " = 1/(4a^2)sin(at){e^(at) - e^(-at)} #

# " " = 1/(4a^2)sin(at)2sinh(at) #

# " " = (sin(at)sinh(at))/(2a^2) #

**Part (B)**

In a similar way to the first solution we want to make the denominator the difference of two squares so that we have two factors leading to two partial fractions. If we complete the square on this denominator the method fails as we get

# s^4+s^2+1 = (s^2+1/2)^2 +3/4#

Which is not the difference of two squares; so let's try a little manipulation:

# s/(s^4+s^2+1) = s/{ (s^4+2s^2+1} -s^2 } #

# " " = s/{ (s^2+1)^2 -s^2 } #

# " " = s/{ (s^2+1-s)(s^2+1+s) } #

Again we find the partial fraction decomposition; which will be of the form;:

# s/(s^4+s^2+1) = A/(s^2+1-s) + B/(s^2+1+s) #

# " " = (A(s^2+1+s) + B(s^2+1-s)) /{ (s^2+1-s)(s^2+1+s) } #

# :. s = A(s^2+1+s) + B(s^2+1-s) #

Equating Coefficients we get:

# Coeff(s) \ \ => 1 = A-B \ \ \ \ \ ..... [1]#

# Coeff(s^2) => 0 = A+B \ \ \ \ \ ..... [2]#

Now we solve these simultaneous equations:

# Eq[1] + Eq[2] => 1=2A #

# :. A=1/2, B= -1/2 #

Se we can write the original expression as:

# s/(s^4+s^2+1) = 1/2{1/(s^2+1-s) - 1/(s^2+1+s)} #

# " " = 1/2{1/((s-1/2)^2-(1/2)^2+1) - 1/((s+1/2)^2-(1/2)^2+1)} #

# " " = 1/2{1/((s-1/2)^2-3/4) - 1/((s+1/2)^2-3/4) } #

# " " = 1/2{1/((s-1/2)^2-(sqrt(3)/2)^2) - 1/((s+1/2)^2-(sqrt(3)/2)^2) } #

And we can now take inverse Laplace Transforms using standard results to get:

# â„’^-1 {s/(s^4+s^2+1)} = 1/2{e^(1/2t)sin(sqrt(3)/2)/(sqrt(3)/2)) - e^(-1/2t)sin(sqrt(3)/2)/(sqrt(3)/2))} #

# " " = 1/2*2/sqrt(3)sin(sqrt(3)/2){e^(1/2t) - e^(-1/2t)} #

# " " = 1/sqrt(3)sin(sqrt(3)/2)2sinh(1/2t) #

# " " = 2/sqrt(3)sin(sqrt(3)/2)sinh(1/2t) #

Featured 2 months ago

# 2+pi/4 #

Here is the graph of the two curves. The shaded area,

It is a symmetrical problems so we only need find the shaded area of the RHS of Quadrant

We could find the angle

# r=1+cos 2theta #

# r=1 #

However, intuition is faster, and it looks like angle of intersection in

# theta = pi/4 => r=1+cos 2theta =1+cos(pi/2) = 1 #

Confirming our intuition. So now we have the following:

Where the shaded area repents

We can now start to set up a double integral to calculate this area

#r# sweeps out a ray from#1# to#1+cos(2theta)#

#theta# varies for#0# to#pi/4#

So then:

# 1/4A = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta #

If we evaluate the inner integral first then we get:

# int_1^(1+cos2theta) r \ dr = [1/2r^2]_1^(1+cos2theta) #

# " " = 1/2( (1+cos2theta)^2 - 1^2) #

# " " = 1/2( 1+2cos2theta+cos^2 2theta - 1) #

# " " = 1/2( 2cos2theta+cos^2 2theta) #

# " " = 1/2( 2cos2theta+1/2(cos4theta+1)) #

# " " = cos2theta+1/4cos4theta+1/4 #

And so our double integral becomes:

# 1/4A \ = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta #

# " " = int_0^(pi/4) cos2theta+1/4cos4theta+1/4 \ d theta #

# " " = [1/2sin2theta+1/16sin4theta+1/4theta]_0^(pi/4) #

# " " = (1/2sin(pi/2)+1/16sinpi+1/4pi/4) - (0) #

# " " = 1/2+pi/16 #

# :. A = 2+pi/4 #

# #

**METHOD 2**

If you are not happy with double integrals, then we can evaluate the area using the polar area formula

Here this shaded area is given by:

# A_1 = 1/2 \ int_0^(pi/4) \ (1+cos 2theta)^2 \ d theta #

# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+ cos^2 2theta \ d theta #

# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2(cos4theta+1) \ d theta#

# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2cos4theta+1/2 \ d theta#

# \ \ \ \ = 1/2 [3/2theta+sin2theta+1/8sin4theta]_0^(pi/4)#

# \ \ \ \ = 1/2 ((3pi)/8+1+0)#

# \ \ \ \ = (3pi)/16+1/2#

And this shaded area

is given by:

# A_2 = 1/2 \ int_0^(pi/4) \ (1)^2 \ d theta #

# \ \ \ \ = 1/2 [theta]_0^(pi/4)#

# \ \ \ \ = 1/2 (pi/4-0)#

# \ \ \ \ = pi/8#

And so the total sought area is

# 1/4 A \ = A_1 - A_2 #

# " " = (3pi)/16+1/2 - pi/8 #

# " " = (pi)/16+1/2 #

#:. A = (pi)/4+2 #

Featured 1 month 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 1 month ago

Use

Recall (from trigonometry class) that

# = (sin^2(8x)(1+cos(3x)))/(sin^2(3x)(1+cos(8x))#

We can see that

so we will focus on

We'll use the facts that

# = lim_(xrarr0)64/9 (sin(8x)/(8x))^2/(sin(3x)/(3x))^2#

# = 64/9 * 1/1 = 64/9#

Featured yesterday

The smallest area is

Let us set up the following variables:

# {(x, "Width of poster (cm)"), (y, "Height of poster (cm)"), (A, "Area of poster ("cm^3")") :} #

Then the dimensions of the **printed matter** are:

# {("Width", =x-6-6,=x-12), ("Height", =y-4-4,=y-8), ( :. " Area",=384 ,=(x-12)(y-8)) :} #

So we have;

# 384 =(x-12)(y-8) #

# :. y-8 = 384/(x-12)#

# :. y = 8+384/(x-12)#

# " " = (8(x-12)+384)/(x-12)#

# " " = (8x-96+384)/(x-12)#

# " " = (8x+288)/(x-12) \ \ \ \ \ ..... (star)#

And the total area of the poster is given by:

# A = xy #

# \ \ \ = (x)((8x+288)/(x-12)) \ \ \ # (using (#star# ))

# \ \ \ = (8x^2+288x)/(x-12) #

We want to minimize (hopefully) by finding

# (dA)/dx = { (x-12)(16x+288) - (1)(8x^2+288x) } / (x-12)^2 #

# " " = { 16x^2+288x-192x+3456-8x^2-288x } / (x-12)^2 #

# " " = { 8x^2-192x+3456 } / (x-12)^2 #

At a min or max

# :. { 8x^2-192x+3456 } / (x-12)^2 = 0 #

# :. 8x^2-192x-3456 = 0 #

# :. 8(x^2-24x-432)=0 #

# :. x^2-24x-432 = 0 #

# :. (x+12)(x-36) = 0 #

This equation leads to the two solutions:

# x= -12# , or# x=36#

Obviously

# y = (288+288)/(36-12) \ \ \ \ \ # (using (#star# ))

# \ \ = 576/24 #

# \ \ = 24 #

With these dimensions we have:

# A = 36*24 #

# \ \ \ = 864 #

We should check that this value leads to a minimum (rather than a maximum) area, which we confirm graphically:

graph{(8x^2+288x)/(x-12) [-2, 50, -2000, 2000]}

Featured 1 week ago

Use the parametric formula:

and substitute:

we have:

Factor the denominator:

and use partial fractions decomposition:

Simplify the expression:

Now:

however because of the symmetry of the expression we can ignore the absolute value, as changing

Using now the properties of logarithms:

Undoing the substitution:

Multiply by

Use now the identity:

Further trigonometric semplification is possible, but the answer is getting too long...

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