Recent content by estalniath

I had recently stumbled upon a computer program that can automatically add missing hydrogen atoms to protein structures from files from the protein data bank. (due to the hydrogen atom having only 1 electron, hence making its electron density too low to be detected by X-ray crystallography...

Hello, It can be derived from basic principles. You can start by finding the strain caused by the thermal expansion. The change in length for this part can be found by subbing in the thermal expansion formuale. Hence we have found the strain. Y=stress/strain =Fl/A(dl) where dl= change...

Hello, 1) I think that you'll have to use the relationship Heat gained by copper calorimeter and water=Heat lost by Steam via condensation +Heat lost by 100 degree celcius of water(after condensation) to 50 degree celcius 2) For question 2, I think that you require the concept of thermal...

Thanks for pointing that out Daniel! I guess that I took the "y" there for granted every time I used the characteristic solution to get the y_h

1)I think that you can make fast work on this question by using McClaurin's expansion for e^x, then divide it later by x to find dy/dx (in a summation notation for easy integration later) 2) For the second part, since the initial value is given we can use the fundamental theorem of calculus to...

Hello, Since the solution is in the form y=ASin3x+BCos3x, The original equation must have complex roots 3i and -3i. Thus, a possible solution is d^2y/dx^2+9=0. =)

I think that you made a mistake in your calculations, subbing 3 in gives 4*(9)-48=36-48=-12 =)

1)Firstly we'll let the expression (n!)^(n^-n)=y, then taking ln on both sides give, lny=ln(n!)/n^n. 2) We'll then find the limit of the expression at the RHS using Le Hopital's rule or by observation that n^n increases faster than ln(n!). =) 3) After finding the limit, L, all we have to...

Your characteristic equation should be of the form y=Asin6x+Bcos6x since the the squareroot of -36 is 6i and -6i =)

After subbing the values of d^2y/dx^2,dy/dx and y into the differential equation, the RHS of the equation should be sin2x, then you'll have to do comparing of the coefficients to determine the value of the constants =)

*Ammendments:Trial solution=CxSin2x+DxCos2x

You'll have to find the homogeneous solution first before applying the method of undetermined coefficients. Let (D^2+4)y=r^2+4=0. hence r=2i,-2i Thus y(homogenous)=Acos2x+Bsin2x Therefore, let the trial function be y=AxSin2x+BxCos2x You should get the answer from here. =)

Hello. 1)You'll have to draw a polar curve to help you out with this question. From the drawn polar curve, you'll get the minimum value of r to be 1. (when cos(theta) is negative) 2) Thus it follows that the range for r is 1<=r<=4+3cos(theta) Hence we'll integrate r from 1 to...

I see. Thank you for the advice =)

You'll have to let A=f(tx,ty,tz)=t^kf(x,y,z) Then find dA/dt= df/d(tx)*dx/dt+df/d(ty)*dy/dt+df/d(tz)*dz/dt =df/d(tx)*x+df/d(ty)*y+df/d(tz)*z For the right hand side, we'll get, k*t^(k-1)f(x,y,z) Then put t=1, and we'll get the equation xdf/dx+ydf/dx+zdf/dz=Kf(x,y,z)...