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Featured 1 month ago

Take the derivative of both sides with respect to

Use the product rule on the left side:

Rearrange so all the terms with

Factor a little:

Solve for

Featured 1 month ago

Given:

#k(dV)/dt+V=0#

Rearrange:

#1/V(dV)/dt=-1/k#

Take the integral of both sides:

#int1/VdV=-1/kintdt#

This gives:

#lnV=-t/k+C#

At

#lnV_0=C#

Hence:

#V=V_0e^(-t/k)#

Or:

#t=kln(V_0/V)#

When

#t=kln10#

Given

#t=50ln10#

Featured 1 month ago

Concave:

Convex:

A function is convex if the second derivative is positive and concave where its second derivative is negative. When the second derivative is

So:

Concave:

Convex:

Graph:

graph{y=-2x^3+4x^2+5x-5 [-4, 4, -7.12, 7.12]}

Featured 1 month ago

Use Newton-Raphson method for numerical approximation.

#f(x)=lnx-e^(cosx)#

Take the derivative with respect to

#f'(x)=1/x+sinxe^(cosx)#

Given the condition

#0=1/x+sinxe^(cosx)#

Let

#u(x)=1/x+sinxe^(cosx)#

Plot

graph{1/x+sinxe^(cosx) [-0.5, 10.5, -2, 4]}

By examination, the roots of

Take the derivative with respect to

#u'(x)=-1/x^2+(cosx-sin^2x)e^(cosx)#

The Newton-Raphson approximation equation is:

#x_(n+1)=x_n-(u(x_n))/(u'(x_n))#

Taking

#x_0=4#

#x_1=3.794906369#

#x_2=3.776053960#

#x_3=3.775920734#

#x_4=3.775920727#

#x_5=3.775920727#

Taking

#x_0=6#

#x_1=6.247412504#

#x_2=6.223989036#

#x_3=6.223939693#

#x_4=6.223939692#

#x_5=6.223939692#

Taking

#x_0=10#

#x_1=9.730092594#

#x_2=9.698501451#

#x_3=9.698236088#

#x_4=9.698236070#

#x_5=9.698236070#

Hence

#x=3.775920727, 6.223939692, 9.698236070#

Featured 2 weeks ago

The first derivative is

Let

Rearranging the term of

For differentiating we use the

and the

Factorizing

Factorizing

The end result is

I hope it helps.

Featured 1 week ago

**LEFT-HAND INTEGRAL**

The **left-hand** integral is trivial, as there are a lot of constants:

#(epsilonA_(s,r)sigma)/(rhoVc)int_(0)^(t)dt = (epsilonA_(s,r)sigma)/(rhoVc) cdot t#

**RIGHT-HAND INTEGRAL: PREPARATION**

The **right-hand** integral starts as:

#int_(T_i)^(T) 1/(T_(sur)^4 - T'^4)dT'#

We wrote **difference of quartics**, so one can factor as follows:

#1/(T_(sur)^4 - T'^4) = 1/((T_(sur)^2 + T'^2)(T_(sur)^2 - T'^2))#

#= 1/((T_(sur)^2 + T'^2)(T_(sur) + T')(T_(sur)-T'))#

This becomes a partial fraction decomposition. We propose, at a **constant surrounding temperature**, the following *two* integrands:

#1/((T_(sur)^2 + T'^2)(T_(sur) + T')(T_(sur)-T'))#

#= overbrace((AT' + B)/((T_(sur)^2 + T'^2)))^"Integrand 1" + overbrace(C/(T_(sur) + T') + D/(T_(sur) - T'))^"Integrand 2"#

**RIGHT-HAND INTEGRAL: ACQUIRING SYSTEM OF EQNS**

When getting common denominators, one obtains:

#1 = (AT' + B)(T_(sur)^2 - T'^2) + C(T_(sur)^2 + T'^2)(T_(sur) - T') + D(T_(sur)^2 + T'^2)(T_(sur) + T')#

Next, simplify by distributing and re-factoring.

#1 = color(red)(AT_(sur)^2T') - color(orange)(AT'^3) + color(purple)(BT_(sur)^2) - color(cyan)(BT'^2)#

#+ color(purple)(CT_(sur)^3) - color(red)(CT_(sur)^2T') + color(cyan)(CT_(sur)T'^2) - color(orange)(CT'^3)#

#+ color(purple)(DT_(sur)^3) + color(red)(DT_(sur)^2T') + color(cyan)(DT_(sur)T'^2) + color(orange)(DT'^3)#

This regroups into a polynomial in orders of

#0T'^3 + 0T'^2 + 0T' + 1#

#= color(orange)((-A - C + D))T'^3 + color(cyan)((-B + CT_(sur) + DT_(sur)))T'^2 + color(red)((AT_(sur)^2 - CT_(sur)^2 + DT_(sur)^2))T' + color(purple)((BT_(sur)^2 + CT_(sur)^3 + DT_(sur)^3))#

As a result, we have the following **system of equations** (since

#-A - C + D = 0# #" "" "" "" "" "bb((1))#

#-B + CT_(sur) + DT_(sur) = 0# #" "" "bb((2))#

#A - C + D = 0# #" "" "" "" "" "" "bb((3))#

#BT_(sur)^2 + CT_(sur)^3 + DT_(sur)^3 = 1# #" "bb((4))#

**RIGHT-HAND INTEGRAL: SOLVING SYSTEM OF EQNS**

In solving this we could do the following.

- Since
#-A - C + D = A - C + D# , it follows that#A = -A# , which is only true if#color(green)(A = 0)# . - Therefore, from
#(1)# or#(3)# ,#C = D# , and we can solve for#B# in#(2)# .

#=> B = 2CT_(sur) = 2DT_(sur)#

- To solve for
#C# or#D# , use#(4)# :

#2CT_(sur) cdot T_(sur)^2 + CT_(sur)^3 + CT_(sur)^3 = 1#

#=> color(green)(C = 1/(4T_(sur)^3) = D)#

- Therefore,
#color(green)(B) = 2 cdot 1/(4T_(sur)^3) cdot T_(sur) = color(green)(1/(2T_(sur)^2))# .

**RIGHT-HAND INTEGRAL: SETTING IT UP**

This means we can insert

#1/((T_(sur)^2 + T'^2)(T_(sur) + T')(T_(sur)-T'))#

#= color(green)((1//2T_(sur)^2))/((T_(sur)^2 + T'^2)) + color(green)((1//4T_(sur)^3))/(T_(sur) + T') + color(green)((1//4T_(sur)^3))/(T_(sur) - T')#

#= overbrace(1/2 1/T_(sur)^2 1/((T_(sur)^2 + T'^2)))^"Integrand 1" + overbrace(1/4 1/T_(sur)^3 [1/(T_(sur) + T') - 1/(T_(sur) - T')])^"Integrand 2"#

The integration is now of the following:

#int_(T_i)^(T) 1/(T_(sur)^4 - T'^4)dT'#

#= overbrace(1/2 1/T_(sur)^2 int_(T_i)^(T) 1/(T_(sur)^2 + T'^2)dT')^"Integral 1" + overbrace(1/4 1/T_(sur)^3 int_(T_i)^(T) [1/(T_(sur) + T') - 1/(T_(sur) - T')]dT')^"Integral 2"#

**RIGHT-HAND INTEGRAL: SOLVING IT**

The second integral here is (before evaluating the integral bounds):

#color(highlight)(1/4 1/T_(sur)^3 int[1/(T_(sur) + T') - 1/(T_(sur) - T')]dT')#

#= 1/(4T_(sur)^3) [ln|T_(sur) + T'| - ln|T_(sur) - T'|]#

#= color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T')/(T_(sur) - T')|)#

The first integral here needs to be manipulated more. Factor out a

#color(darkblue)(1/(2T_(sur)^2) int1/(T_(sur)^2 + T'^2)dT')#

#= 1/(2T_(sur)^4) int1/(1 + (T'//T_(sur))^2)dT'#

This resembles the integral of

Thus, **cubed term** on the outside like we expected when we integrate:

#=> 1/(2T_(sur)^3) int1/(1 + u^2)du#

#= 1/(2T_(sur)^3) arctanu#

#= color(darkblue)(1/(2T_(sur)^3) arctan((T')/T_(sur)))#

**FORMING/SIMPLIFYING THE RESULT**

Now we can **combine** the two overarching integrals to get:

#int_(T_i)^(T) 1/(T_(sur)^4 - T'^4)dT'#

#= |[color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T')/(T_(sur) - T')|) + color(darkblue)(1/(2T_(sur)^3) arctan((T')/T_(sur)))]|_(T_i)^(T)#

Evaluating this on the integral bounds, we get:

#= [color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T)/(T_(sur) - T)|) + color(darkblue)(1/(2T_(sur)^3) arctan((T)/T_(sur)))] - [color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T_i)/(T_(sur) - T_i)|) + color(darkblue)(1/(2T_(sur)^3) arctan((T_i)/T_(sur)))]#

Regroup terms to get:

#= color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T)/(T_(sur) - T)| - 1/(4T_(sur)^3) ln|(T_(sur) + T_i)/(T_(sur) - T_i)|) + color(darkblue)(1/(2T_(sur)^3) arctan((T)/T_(sur)) - 1/(2T_(sur)^3) arctan((T_i)/T_(sur)))#

To factor this, multiply the third and fourth terms by

#= 1/(4T_(sur)^3) {color(highlight)(ln|(T_(sur) + T)/(T_(sur) - T)| - ln|(T_(sur) + T_i)/(T_(sur) - T_i)|) + color(darkblue)(2[arctan((T)/T_(sur)) - arctan((T_i)/T_(sur))])}#

**FINALIZING THE RESULT**

Now, we can equate this entire **right**-hand integral with the **left**-hand result we got on the first few lines:

#(epsilonA_(s,r)sigma)/(rhoVc) cdot t#

#= 1/(4T_(sur)^3) {ln|(T_(sur) + T)/(T_(sur) - T)| - ln|(T_(sur) + T_i)/(T_(sur) - T_i)| + 2[arctan((T)/T_(sur)) - arctan((T_i)/T_(sur))]}#

Multiply the left-hand constants over to get:

#color(blue)(t = (rhoVc)/(4epsilonA_(s,r)sigmaT_(sur)^3) {ln|(T_(sur) + T)/(T_(sur) - T)| - ln|(T_(sur) + T_i)/(T_(sur) - T_i)| + 2[arctan((T)/T_(sur)) - arctan((T_i)/T_(sur))]})#

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