Finding Differential Equations Calculus homework?
The function y satisfies a differential equation of the form y' = ky for some number k. If you are told that when t = 3 that y is 5 and the rate of change of y is 3 then what is k? I know that K= 3/5, but the second part of the question asks, What is y(6)? y(6)=?
Help with homework? Finding differential in Calc III. More inside!?
So the question goes as such: 'The gas equation for one mole of oxygen relates its pressure, P (in atmospheres), its temperature, T (in K), and its volume, V (in cubic decimeters, dm3): T=16.574*1/V - 0.52754*1/V^2 - 0.3879P + 12.187VP a) Find the temperature T and differential dT if the volume is 32 dm3 and the pressure is 1.5 atmosphere. T = ? dT = ? b) use your answer to part (a) to estimate how much volume would have to change if the pressure increased to 0.1 atmosphere and the temperature remained constant. change in volume = ?' I found the temperature, T, to be 584.9115723. My differentiable, dT, was wrong; apparently, it was not (-0.016)dV + (389.5961)dP. Part b, I'm simply clueless. Any help is appreciated. Thanks!
Calculus 2 homework help?
Any equation of the form dS/dt = k (S - M)^2 with k and M constants, models this situation. (M is the maximal amount, and k is the constant of proportionality between dS/dt and (S - M)^2). This is a separable differential equation. Using the standard method for solving it, you can write it as dS/(S - M)^2 = k dt, and then integrate both sides to learn learn that -1/(S - M) = kt + C for some constant C, which implies that -1/(kt + C) = S - M, which implies that S = M - 1/(kt + C) for some constant C. If we take k = 3 and M = 2 then we have S(t) = 2 - 1/(3t + C) and since S(0) = 1 we must have 1 = 2 - 1/(3*0 + C), which implies that 1 = 2 - 1/C, which implies that 1/C = 1, or that C = 1, so that S(t) = 2 - 1/(3t + 1) for all t.
AP Calculus HomeWork help!!?
As you don't include the form of the model that uses the parameters mentioned, I'll use the one given on wikipedia. dP/dt = r*P*(1 - (P/k)) r is the growth rate, k is the carrying capacity (a) In our case... dP/dt = 0.06P - 0.000075P^2 dP/dt = 0.06*P*(1 - 0.00125P) dP/dt = 0.06*P*(1 - P/800) Equating term by term... the growth rate (r) in our case is 0.06 the carrying capacity (k) in our case is 800 (b) To find a formula for the population we need to integrate to solve for P in terms of t. dP/dt = 0.06*P*(1 - P/800) We isolate everything involving P (dP and polynomial of P) dP/ ( 0.06*P*(1 - P/800) ) = dt Int( dP / ( 0.06*P*(1 - P/800) ) ) = Int( dt ) = t To do the integral on the left we use partial fractions to split up the integral. Int( dP / ( 0.06*P*(1 - P/800) ) ) = (50/3)*Int( 800*dP / ( P*(800-P) ) ) = (50/3)*( Int( dP/P ) + Int( dP/(800-P) ) ) = (50/3)*( Ln(P) - Ln(800-P) ) = t + C Ln(P/(800-P)) = (3/50)*t + C Since multiplying a constant by a constant leaves a constant we keep calling it C even though it is different from the old C P/(800-P) = C*e^((3/50)*t) P = 800*C*e^((3/50)*t) / (1 + C*e^((3/50)*t) ) We can't lump in the 800 because C is in more than one place. Using our initial condition we solve for C P(0) = 25 = 800*C / (1+C) 25+25*C = 800*C 25 = 775*C C = 1/31 So our equation for population in terms of time is... P = 800*e^((3/50)*t) / ( 31 + e^((3/50)*t) ) (c) 450 = 800*e^((3/50)*t) / ( 31 + e^((3/50)*t) ) 31*450 = e^((3/50)*t) * (800 - 450) 13950/350 = e^((3/50)*t) 279/7 = e^((3/50)*t) Ln(279/7) = (3/50)*t (50/3) * Ln(279/7) = t t is approximately 61.4 years
Calculus Homework Help?
One side of a right triangle is known to be 45 cm long and the opposite angle is measured as 30°, with a possible error of 1°. (a) Use differentials to estimate the error in computing the length of the hypotenuse. (Round the answer to two decimal places.) (b) What is the percentage error? (Round the answer to the nearest whole number.) i got .245 for part a but it was wrong :(
More AP Calculus Integration Homework Help!?
It's a separable differential equation, so if you "multiply" both sides by dx and e^(2y) and integrate, you get: ∫ e^(2y) dy = ∫ 3x² dx Both sides of this are straightforward to integrate: 1/2 e^(2y) = x³ + C Multiplying both sides by 2: e^(2y) = 2x³ + C Take the natural log of both sides: 2y = ln(2x³ + C) And divide by 2 to solve for y: f(x) = y = 1/2 ln(2x³ + C) Then use the initial conditions to find C: f(0) = 1/2 = 1/2 ln(2(0)³ + C) 1/2 = 1/2 ln(C) 1 = ln(C) and then raising both sides to the e power: e = C So the function that satisfies the equation and initial condition should be: y = f(x) = 1/2 ln(2x³ + e) Just to double check, I'll differentiate: y' = 1/2 (1/(2x³ + e)) (3x²) = 3x²/(2(2x³ + e)) but e^(2y) = e^(2(1/2) ln(2x³ + e)) = 2x³ + e Somewhere there's a little mistake, but I can't find it, cuz, I end up getting: y' = 3x²/(2e^(2y)) Which is wrong by a factor of 2 in the denominator...but I can't find my mistake...maybe you can. b) If the correct function is: f(x) = 1/2 ln(2x³ + e) Then, for the domain, you need to ensure that the argument to the natural log does become zero or negative: 2x³ + e > 0 2x³ > -e x³ > -e/2 x > ³√(-e/2) So it appears the domain is (³√(-e/2), ∞). The range of the function is the same as the range of natural log, (-∞, ∞).
What area should I work if I like calculus, differential equations, programming, economy and languages?
Finance, definitely.Physics, applied mathematics, chemistry, mechanical/chemical/electrical engineering also very good.The pure sciences won’t have much contact with economics, but certain parts of engineering will.Languages exposure is limited in all cases to reading foreign publications, but Russians and French publications are particularly strong in mathematics, and parts of physics.
Calculus homework help, please?
1. Use differentials to estimate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter 40 meters. (Recall that the volume of a sphere of radius r is V = (4/3)πr^3. Notice that you are given that dr = 0.02.) ⇒ 2. 1. For a cylinder with surface area 50, including the top and the bottom, find the ratio of height to base radius that maximizes the volume.
Calculus Homework?
An egg of a particular bird is very nearly spherical. The radius to the inside of the shell is 88 millimeters and the radius to the outside of the shell is 8.28.2 millimeters. Use differentials to approximate the volume of the shell. [Remember that Upper V left parenthesis r right parenthesis equals four thirds pi r cubedV(r)= 4 3π r3, where r is the radius.]