Title 90 Applications of DE.pdf ✅

MSTC 89: Applications of Differential Equations 1. Exponential Growth and Decay The rate of change of a quantity is its first derivative. That quantity grows (or increases) when the first derivative is positive. The quantity decays (or decreases) when the first derivative is negative. When the first derivative is directly proportional to the original quantity, dA dt = kA the function representing the quantity is exponential. This equation is variably separable. Let A0 be the original quantity. dA A = k dt ∫ dA A At A0 = ∫ k dt T 0 ln At − ln A0 = kT ln At A0 = kT At A0 = e kT At = A0e kT The doubling time is the time it takes for an object to grow to twice the original size. The half-life is the time it takes for an object to decay to half the original size. Since this function is exponential, it can be modeled using the exponential function in the calculator (MODE- EQN-ab EXP or MODE-EQN-e exp) The time is entered in the x-column, and the quantity is entered in the y-column. When using exponential modeling, two pairs of data are needed. One of the pairs commonly used is the initial value at t = 0. Exponential modeling does not work when the first derivative is not directly proportional to the quantity only (for example, the derivative is proportional to the “square” of the quantity). 2. Newton’s Law of Cooling For an object with initial temperature T0 in surroundings with temperature TS , dU dt = k(U − Ts ) Separating the variables, dU U − TS = k dt ∫ dU U − TS T T0 = ∫ k dt t 0 ln(T − Ts ) − ln(T0 − Ts ) = kt T − Ts T0 − Ts = e kt
This means that cooling bodies can also be modeled using exponential functions. The time is entered in the x-column. The difference between the temperature of the object and the surroundings is entered in the y- column. 3. Kinematics (See MSTC 100: Rectilinear Motion) The motion of objects with uniform motion can be modeled using a linear function. s = so + vt where s = the position at any time, so = initial position, v = velocity, t = time Use linear models, s = so + vt → y = A + Bx By correspondence, y = s, A = s0, B = v, x = t. The motion of objects with uniform acceleration can be modeled using a quadratic function. s = so + vt + 1 2 at 2 where s = the position at any time, so = initial position, v = velocity, a = acceleration, t = time Use quadratic models, s = so + vt + 1 2 at 2 → y = A + Bx + Cx 2 By correspondence, y = s, A = s0, B = v, C = 1 2 a, x = t. 4. Terminal Velocity For falling objects with mass m, pulled by gravity g, and affected by air resistance, the air resistance is proportional to the velocity. m ⏞ a dv dt = mg − kv dv dt + k m v = g Solving this equation as a linear differential equation, ve k m t = mg k e k m t + C v = mg k + Ce − k m t T0 − Ts T − Ts t 0
When t = 0, 0 = mg k + C C = − mg k When t → ∞, v = vt , vt = mg k + C(0) k = mg vt Therefore, v = vt (1 − e − gt vt) 5. Escape Velocity For a celestial body with radius R and acceleration due to gravity g, the required velocity to escape its gravitational pull is called the escape velocity. ve = √2gR 6. Mixing Problems Consider a tank containing a mixture with volume V0 and concentration c0. From mixture problems in algebra, A0 = c0V0 When a mixture with different concentrations is introduced at a rate of vin, then the change in the amount of solute in the mixture is represented by the differential equation dA dt = cinvin When the uniformly mixed solution also flows out of the container at a rate vout, then the differential equation becomes dA dt = cinvin − coutvout where the concentration of the outgoing mixture is cout = A V0 + (vin − vout)t The third equation is solved as a linear differential equation. 7. Undamped Vibrations For a spring with mass m and spring constant k, ma = −kx ma + kx = 0 a + k m x = 0