Featured Answers

11
Active contributors today

Answer:

#(dy)/dx=cos(y)/(1+x*sin(y))#

Explanation:

Take the derivative of both sides with respect to #x#:

#d/dx(x*cos(y))=d/dx(y)#

Use the product rule on the left side:

#x*-sin(y)(dy)/dx+cos(y) = (dy)/dx#

Rearrange so all the terms with #(dy)/dx# are on the same side:

#(dy)/dx+x*sin(y)(dy)/dx=cos(y)#

Factor a little:

#(dy)/dx(1+x*sin(y))=cos(y)#

Solve for #(dy)/dx#:

#(dy)/dx=cos(y)/(1+x*sin(y))#

Answer:

#t=50ln10#

Explanation:

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 #t=0#, #V=V_0#:

#lnV_0=C#

Hence:

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

Or:

#t=kln(V_0/V)#

When #V=1/10V_0#:

#t=kln10#

Given #k=50#:

#t=50ln10#

Answer:

Concave: #{x in RR: 2/3< x < oo}#

Convex: #{x in RR: -oo < x < 2/3 }#

Explanation:

A function is convex if the second derivative is positive and concave where its second derivative is negative. When the second derivative is #0# this could mean the function is concave , convex, or it could be a point of inflexion. This would have to be tested using the first derivative.

#f(x)=-2x^3+4x^2+5x-5#

#dy/dx(-2x^3+4x^2+5x-5)=-6x^2+8x+5#

#(d^2y)/(dx^2)(-6x^2+8x+5)=-12x+8#

So:

#-12x+8<0#

#x>8/12#

#x >2/3#

Concave: #{x in RR: 2/3< x < oo}#

#-12x+8>0#

#x<8/12#

#x<2/3#

Convex: #{x in RR: -oo < x < 2/3 }#

Graph:

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

Answer:

Use Newton-Raphson method for numerical approximation.

Explanation:

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

Take the derivative with respect to #x#:

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

Given the condition #f'(x)=0#:

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

Let

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

Plot #u(x)# over #0< x<=10#:
graph{1/x+sinxe^(cosx) [-0.5, 10.5, -2, 4]}

By examination, the roots of #u(x)# are close to #4#, #6#, and #10#.

Take the derivative with respect to #x#:

#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# gives:

#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# gives:

#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# gives:

#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#

Answer:

The first derivative is #quad (-48x * (x^2+4)^2)/((x^2-4)^4)#.

Explanation:

Let #quad f(x) = ( (x^2+4)/(x^2-4) )^3#.

Rearranging the term of #f# gives us

#quad f(x) = ((x^2+4)/(x^2-4))^3 = (x^2+4)^3/(x^2-4)^3#.

For differentiating we use the #color(green)("chain rule")# for #color(blue)(u^'(x))# and #color(blue)(v^'(x))#

#color(green)(d/dx (u(v(x))) = (u(v(x)))^' = u^'(v(x)) * v^'(x) )# (outside times inside derivative)

and the #color(blue)("quotient rule")# for the whole term of #f#

#color(blue)( d/dx (u(x))/( v(x)) = ((u(x))/(v(x)))^' = (u^'(x) * v(x) - u(x) * v^'(x))/(v(x))^2)# (with #v(x) != 0#).

#f^'(x) = #

#( 3 * 2x * (x^2+4)^2 * (x^2-4)^3 - (x^2+4)^3 * 3 * 2x * (x^2-4)^2 )/( x^2-4)^6 #.

Factorizing #(x^2-4)^2# from both summands of the enumerator gives us

#f^'(x) = #

#( cancel((x^2-4)^2) * [ 6x * (x^2+4)^2 * (x^2-4)- 6x * (x^2+4)^3 ] )/( x^2-4)^cancel(6^4) #.

Factorizing #6x * (x^2+4)^2# from both summands of the enumerator leads to

#f^'(x) = ( 6x * (x^2+4)^2 * [ x^2-4- (x^2+4)] )/( x^2-4)^4 #

#f^'(x) = ( 6x * (x^2+4)^2 * (- 8) )/( x^2-4)^4 #.

The end result is

#color(red)(f^'(x) = ( -48x * (x^2+4)^2 )/(x^2-4)^4)#.

I hope it helps.

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 #T'# here because it need not be equal to #T#. This is clearly a 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 #T'# up to #3#:

#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 #T_(sur) ne 0#, we divided it out of #(3)#):

#-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 #A#, #B#, #C#, and #D# to get:

#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 #T_(sur)^2# to get:

#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 #arctanu#. Let #u = T'//T_(sur)#, so that #du = dT'//T_(sur)#.

Thus, #T_(sur)du = dT'#, and that then gives us the 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 #2/2#, then factor out #1/(4T_(sur)^3)# to get:

#= 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))]})#

View more
Questions
Ask a question Filters
Loading...
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions