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For a non-rigorous proof, please see below.

For a positive central angle of

Source:

The geometric idea is that

So we have:

For small positive

So

so

We also have, for these small

so

Since both one sided limits are

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There are three possible coordinates:

# ( 1, 6 ) ; ( -1-1/(3sqrt(2)), 13/6+(2sqrt(2))/3 ) ; ( -1+1/(3sqrt(2)), 13/6-(2sqrt(2))/3 ) #

If we choose

# 8y + x - 49 = 0 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is

We have:

# f(x) = 3x^2+2x+1 #

Differentiating wrt

# f'(x) = 6x+2 #

Suppose the point we seek has coordinates

# b = 3a^2+2a+1 # ..... [A]

The slope of the tangent line at

# m_T = f'(a) = 6a+2 #

So, the slope of the normal line is given by:

# m_N = -1/(6a+2) #

As the normal also passes through

# y - 5 = -1/(6a+2)(x-9) # ..... [B]

This normal equation also passes through the point

# b - 5 = -1/(6a+2)(a-9) #

Substituting [A] gives us:

# 3a^2+2a+1 - 5 = -(a-9)/(6a+2) #

# :. (6a+2)(3a^2+2a- 4) = -a+9 #

# :. 18a^3+12a^2- 24a + 6a^2+4a- 8 = -a+9 #

# :. 18a^3+18a^2- 19a -17 = 0 #

The challenge is to now solve this cubic for

# (a-1)(18a^2+36a+17) = 0 #

Then using the quadratic formula, we gain the additional roots, and so we have:

# a=1, -1+-1/(3sqrt(2)) #

Using [A] with these values of

# a= { (1), (-1-1/(3sqrt(2))), (-1+1/(3sqrt(2))) :} => b= { (6), (13/6+(2sqrt(2))/3), (13/6-(2sqrt(2))/3) :}#

Hence there are three possible coordinates:

# ( 1, 6 ) #

# ( -1-1/(3sqrt(2)), 13/6+(2sqrt(2))/3 ) #

# ( -1+1/(3sqrt(2)), 13/6-(2sqrt(2))/3 ) #

If we choose

# y - 5 = -1/(6+2)(x-9) #

# :. 8y - 40 = -x+9 #

# :. 8y + x - 49 = 0 #

We can confirm this graphically:

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# f_1(t) = â„’^(-1){ 1/(s^2+4s+13)^2 } #

# \ \ \ \ \ \ \ = 1/54 \ e^(-2t) { sin(3t) - 3tcos(3t) } #

# f_2(t) = â„’^(-1){ 1/( (s^2+4)(s+1)^2 } #

# \ \ \ \ \ \ \ = e^(-t)/50 {10t+4 -3sin2t - 4cos2t} #

We have:

A:

# F_1(s) = 1/(s^2+4s+13)^2 #

B:# F_2(s ) =1/( (s^2+4)(s+1)^2 ) #

The Laplace Convolution Theorem tells us that if we define the convolution of two function

# (f star g)(t) = int_0^t \ f(t-x) \ g(x) \ dx #

Then:

# â„’ \ { (f star g)(t)} = F(s)G(s) iff â„’^(-1) \ { F(s)G(s) }= (f star g)(t) #

We will need the following standard Laplace transform and inverses:

# {: (ul(f(t)=â„’^(-1){F(s)}), ul(F(s)=â„’{f(t)}), ul("Notes")), (f(t), F(s),), (t^n, (n!)/(s^(n+1)),n in NN), (e^(at), 1/(s-a), a " constant"), (t^n e^(at), (n!)/(s-1)^(n+1), a " constant," nin NN), (sinat, a/(s^2+a^2), a " constant"), (e^(at)sinbt, b/((s-a)^2+b^2),a","b " constant") :} #

======================================================

**Part [A]:**

We seek:

# f_1(t) = â„’^(-1){ 1/(s^2+4s+13)^2 } = â„’^(-1){ F(s)G(s) }#

Where:

# F(s) = G(s) = 1/(s^2+4s+13) #

Completing the square, we can write as follows:

# F(s) = 1/((s+2)^2-2^2+13) #

# \ \ \ \ \ \ \ = 1/((s+2)^2 + 9) #

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

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

Then using the above table of transforms, we get:

# f(t) =g(t) = 1/3 e^(-2t)sin3t #

So applying the convolution theorem, we have:

# f_1(t) = (f star g)(t) = int_0^t \ f(t-x) \ g(x) \ dx #

# \ \ \ \ \ \ \ \ = int_0^t \ 1/3 e^(-2(t-x))sin3(t-x) \ 1/3 e^(-2x)sin(3x) \ dx #

# \ \ \ \ \ \ \ \ = 1/9 \ int_0^t \ e^(-2t+2x-2x) \ sin(3t-3x) \ sin(3x) \ dx #

Now we apply the trigonometric identity:

# 2sinAsinB = cos(A - B) - cos(A + B) #

Then we have:

# f_1 = 1/9 e^(-2t) \ int_0^t \ (cos(3t-6x) - cos(3t))/2 \ dx #

# \ \ = 1/18 e^(-2t) [ sin(6x-3t)/6 - xcos(3t) ]_0^t #

# \ \ = 1/18 e^(-2t) { (sin(6t-3t)/6 - tcos(3t)) - (sin(-3t)/6 - 0) } #

# \ \ = 1/18 e^(-2t) { sin(3t)/6 - tcos(3t) - sin(-3t)/6 } #

# \ \ = 1/18 e^(-2t) { sin(3t)/6 - tcos(3t) + sin(3t)/6 } #

# \ \ = 1/18 e^(-2t) { sin(3t)/3 - tcos(3t) } #

# \ \ = 1/18 * 1/3 \ e^(-2t) { sin(3t) - 3tcos(3t) } #

# \ \ = 1/54 \ e^(-2t) { sin(3t) - 3tcos(3t) } #

======================================================

**Part [B]:**

We seek:

# f_2(t) = â„’^(-1){ 1/( (s^2+4)(s+1)^2 ) } = â„’^(-1){ F(s)G(s) }#

Where:

#F(s) = 1/(s+1)^2 # ; and# G(s) = 1/(s^2+4) #

Then using the above table of transforms, we get:

# F(s) = 1/(s+1)^2 \ \ \ => f(t) = te^(-t) #

# G(s) = 1/2 \ 2/(s^2+2^2) => g(t) = 1/2sin(2t) #

So applying the convolution theorem, we have:

# f_2(t) = (f star g)(t) = int_0^t \ f(t-x) \ g(x) \ dx #

# \ \ \ \ \ \ \ = int_0^t \ (t-x)e^(-(t-x)) \ 1/2 \ sin(2x) \ dx #

# \ \ \ \ \ \ \ = 1/2 \ int_0^t \ (t-x)e^(x-t) \ sin(2x) \ dx #

Due to the length of the solution thus far, I will omit the derivation of the integral, and quote the result:

# f_2 = 1/2 [-e^(x-t) { ((5x-5t+3)sin2x)/25 + ((-10x+10t+4)cos2x)/25}]_0^t #

# \ \ = [-e^(x-t)/50 { (5x-5t+3)sin2x + (-10x+10t+4)cos2x}]_0^t #

# \ \ = -e^(0)/50 { 3sin2t + 4cos2t} + e^(-t)/50 { (-5t+3)sin0 + (10t+4)cos0} #

# \ \ = e^(-t)/50 {10t+4 -3sin2t - 4cos2t} #

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# int_0^(pi/2) \ sinx \ dx = 1 #

By definition of an integral, then

# int_a^b \ f(x) \ dx #

represents the area under the curve

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#

And we partition the interval

# Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } #

# \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b } #

Here we have

# Delta = {0, 0+1(pi/2)/n, 0+2 (pi/2)/n, 0+3 (pi/2)/n, ..., pi/2 } #

And so:

# I = int_0^(pi/2) \ sinx \ dx #

# \ \ = lim_(n rarr oo) (pi/2)/n sum_(i=1)^n \ f(0+i*(pi/2)/n)#

# \ \ = lim_(n rarr oo) pi/(2n) sum_(i=1)^n \ sin((ipi)/(2n) )# ..... [A}

If we were dealing with polynomials, we would now utilise standard summation formulas for powers, but as we are dealing with a sine summation those results will not help.

In order to evaluate the sine sum, We consider the following:

# 1 + z + z^2 + ... + z^n = (z^(n+1)-1)/(z-1) \ \ # for z#!= 1#

in combination with Euler's formula by taking

#z = e^(i theta) =cos theta+i sin theta#

Then applying De Moivre's theorem:

# e^(i n theta)=cosntheta+isin n theta #

and equating real and imaginary parts, we eventually find that:

# sum_(j=1)^n sin(jtheta) = (cos(theta/2)-cos(n+1/2)theta) / (2sin(theta/2)#

# " " = (cos(theta/2)-cos(ntheta+theta/2)) / (2sin(theta/2)#

# " " = (cos(theta/2)-(cos ntheta cos(theta/2) - sinnthetasin(theta/2))) / (2sin(theta/2)#

# " " = (cos(theta/2)-cos ntheta cos(theta/2) + sinnthetasin(theta/2)) / (2sin(theta/2))#

So, if we put

# I = lim_(n rarr oo) pi/(2n) {(cos((pi)/(4n))-cos (pi/2) cos((pi)/(4n)) + sin (pi/2) sin((pi)/(4n))) / (2sin(pi/(4n)))}#

# \ \ = lim_(n rarr oo) pi/(2n) {(cos((pi)/(4n)) + sin((pi)/(4n))) / (2sin(pi/(4n)))}#

# \ \ = pi/4 \ lim_(n rarr oo) 1/(n) (cot((pi)/(4n)) + 1 ) #

# \ \ = pi/4 { lim_(n rarr oo) 1/(n) (1/tan((pi)/(4n))) + lim_(n rarr oo) 1/(n) }#

# \ \ = pi/4 { lim_(n rarr oo) 1/(n) (( pi/(4n))/tan((pi)/(4n))) * (4n)/pi + 0 }#

# \ \ = lim_(n rarr oo) ( pi/(4n))/tan((pi)/(4n)) #

And a standard calculus limit is:

# lim_(x rarr 0) x/tanx =1 #

Utilising this result with

# I = 1 #

**Using Calculus**

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

# int_0^(pi/2) \ sinx \ dx = [ -cos ]_0^(pi/2) #

# " " = -(cos (pi/2)-cos0) #

# " " = -(0-1) #

# " " = 1 #

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The answer is

We need

The function is

We differentiate with respect to x

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We are given;

The first states that

If a function is increasing, then we can infer that its derivative function must be positive. This can be written as;

In order to show that

With the information we have, only one term is potentially negative,

Of course, we already made the assumption that

Lastly,

The numerator

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The average rate of change is the slope of the secant between:

#(x, g(x))# and#(x+h, g(x+h))#

The slope

#m = (y_2-y_1)/(x_2-x_1)#

So the slope of the secant is:

#(g(x+h)-g(x))/((x+h)-x) = (g(x+h)-g(x))/h#

So given:

#g(x) = 1/(x-2)#

its average rate of change over

#(1/(color(blue)((x+h))-2)-1/(color(blue)(x)-2))/h = ((color(red)(cancel(color(black)(x)))-color(brown)(cancel(color(black)(2))))-(color(red)(cancel(color(black)(x)))+h-color(brown)(cancel(color(black)(2)))))/(h(x-2)(x+h-2))#

#color(white)((1/(((x+h))-2)-1/((x)-2))/h) = (-color(red)(cancel(color(black)(h))))/(color(red)(cancel(color(black)(h)))(x-2)(x+h-2))#

#color(white)((1/(((x+h))-2)-1/((x)-2))/h) = -1/((x-2)(x-2+h))#

In particular, we find:

#lim_(h->0) (g(x+h)-g(x))/h = -1/(x-2)^2#

This is the slope of