Recent content by Vandella

Thanks for your help will just leave it now and not confuse myself even more

Sorry when I said solve I meant evaluate. Could I switch the order of integration so when I evaluate after integrating a second time I lose x and y from the equation?

Ok here goes u= x^2 that gives ∫ e^(-u) integration by substitution gives me -(e^x^2)/2 Possibly :) Now do I just solve for x=1 and x=y?

Well I started by integrating with respect to y and got ye^(-x^2)dx,I then solved for the values y=x and y=0 leaving xe^(-x^2)dx and unsure of if my first steps are correct and how to continue. My idea for a next step would be integration by substitution u=-x^2 but unsure as to whether that...

Homework Statement Please help I need to find the area of e^(-x^2) with the domain 0 is less than or equal to y which is less than or equal to 1 and y is less than or equal to x which is less than or equal to one Homework Equations The Attempt at a Solution I set up the...

oops was not as hard as i first thought y=8-4x

Homework Statement Find an equation of the line passing through the point (1, 4) such that the area of the triangle formed by this line and the positive coordinate axes is as small as possible. Homework Equations The Attempt at a Solution No idea where to start this one please...

Homework Statement A 1.00 mol sample of an ideal monatomic gas is taken through the cycle shown in the figure. The process A B is a reversible isothermal expansion where PA = 5.0 atm, PB = 3.0 atm, VA = 30.0 L, and VB = 50.0 L. (a) Calculate the net work done by the gas. (b)...

ahhh i can't believe i missed that thanks

Homework Statement Find the point(s) on the graph of the function at which the tangent line has the indicated slope. (If an answer does not exist, enter DNE.) g(x) = (1/3)x^3 - (1/2)x^2 - 4x +8 mtan=-4 Homework Equations [b]3. The Attempt at a Solution firstly i derived...

ahh i give up for one night :) 3 hours on one question is longer than i spent completing all the previous 6 questions on the assignment. will look over it again at some point during the weekend. Dont make plans to keep busy :) thanks once again

Calculus i didnt find too bad, just don't like it :) mechanics is a problem didnt do any at A level Thanks for all your help tonight

part a) f=mg-mkv^2 = ma =(this is where i may go wrong) m(dv/dx)(dx/dt) =m(dv/dx)v limited knowledge of differential equations but here goes x = -∫(v/(kv² + g))dv = -(1/(2k))ln(kv² + g) + C. final velocity = initial velocity (v=u) when x = 0, C = (1/(2k))ln(ku² + g). Therefore -2kx =...

no integration needed is the best news i have heard for a while :) i have looked at terminal velocity and think i am getting somewhere with that terminal velocity is reached when force due to gravity = air resistance mg=kv^2 air resistance can be written as 1/2 (drag coefficient x density x...

arghh more integration :( so acceleration = dv/dt and then substitute into v-u=ah ? ohh and thanks a lot for all your help