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UNIT 1 INDIVIDUAL PROJECTTrig Calculus 1) Calculate the derivative dy/dx of sin(2x + 1).a) sin(2x + 1) b) 2cos(2x + 1) c) cos(2x + 1) d) ? cos(2x + 1) 2) Calculate the derivative dy/dx of sin(x2 + 1).a) cos(2x) b) 2cos(x2 + 1) c) (2x) cos(x2 + 1) d) ? cos(x2 + 1) 3) Calculate the derivative dy/dx of sin2(x + 1).a) 2cos(x + 1) b) cos2(x + 1) c) (2x) cos(x + 1) d) 2sin(x + 1)cos(x + 1) 4) Calculate ∫ sin(2x + 1) dx.a) sin(2x + 1) + Cb) ? sin(2x + 1) + C c) cos(2x + 1) + Cd) ? cos(2x + 1) + C 5) Calculate ∫π2π sin(2x + 1) dx.a) πb) ? c) 2πd) ?π422570526268y = logbx means by = xlog(A × B) = log(A) + log(B)log(A / B) = log(A) – log(B)log(AB) = B × log(A) logbM = logaM / logab00y = logbx means by = xlog(A × B) = log(A) + log(B)log(A / B) = log(A) – log(B)log(AB) = B × log(A) logbM = logaM / logabExponential and Logarithmic Equations 6) Calculate the value of x: log5 x = 3.a) 5 b) 25 c) 125 d) 15 7) Rewrite the following as an exponential equation: loga 10 = 16.a) a16 = 10 b) 1016 = a c) 1610 = a d) a10 = 16 8) Rewrite the following as a logarithmic equation: x = b2y.a) logx b = 2y b) logx 2y = b c) logb x = 2y d) logb 2y = xCalculus of Exponents and Logs 9) Calculate ∫ (2ex – 8x) dx.a) 2ex – 8x/ln(8) + cb) 2exln(2) – 8x + c c) 2ex – 8x ln(8) + c d) 2exln(x) – 8x ln(8) + c 10) Calculate ∫ (ln(x)) dx.(Assuming a complex-valued logarithm)a) ln(x) – 1 + cb) ln(x – 1) + c c) x ln(x) + c d) x (ln(x) – 1) + c-4267204365918Useful Engineering Formulas v(t) = dω/dqp(t) = dω/dtv(t) is the voltage (in volts). = (dω/dq)(dq/dt)ω is the energy (in joules). = v(t) × i(t)q is the charge (in coulombs).p(t) is the power in joules/seconds or watts.i(t) = dq/dtvcoil = N dw/dti(t) is the current (in amperes).vcoil is the instantaneous coil voltage (in volts).t is the time (in seconds).N is the number of primary turns.ic = C dv/dt VL = L di/dtic is instantaneous capacitor VL is the instantaneous inductor voltage.current in amperes.L is the inductance in henrys (H).C is capacitance in farads (F).The DC at a point in a circuit is defined to be the total number of coulombs (electricity conveyed in 1 second by a current of 1 ampere) divided by the time.The mathematical model relating charge to voltage in a capacitor is as follows:Δq = CΔvC is a constant capacitance value (in farads).v is voltage (in volts).q is charge (in coulombs).Work is one of the concepts defined differently in science and engineering from the rest of the world.Work = Force × DistanceWork is measured in joules (J). Energy is also measured in joules. Power is the ratio of total work done to the total time taken: Pavg = Δw/Δt.Power is measured in watts.Power is the time derivative of work: P = dW/dt. 0Useful Engineering Formulas v(t) = dω/dqp(t) = dω/dtv(t) is the voltage (in volts). = (dω/dq)(dq/dt)ω is the energy (in joules). = v(t) × i(t)q is the charge (in coulombs).p(t) is the power in joules/seconds or watts.i(t) = dq/dtvcoil = N dw/dti(t) is the current (in amperes).vcoil is the instantaneous coil voltage (in volts).t is the time (in seconds).N is the number of primary turns.ic = C dv/dt VL = L di/dtic is instantaneous capacitor VL is the instantaneous inductor voltage.current in amperes.L is the inductance in henrys (H).C is capacitance in farads (F).The DC at a point in a circuit is defined to be the total number of coulombs (electricity conveyed in 1 second by a current of 1 ampere) divided by the time.The mathematical model relating charge to voltage in a capacitor is as follows:Δq = CΔvC is a constant capacitance value (in farads).v is voltage (in volts).q is charge (in coulombs).Work is one of the concepts defined differently in science and engineering from the rest of the world.Work = Force × DistanceWork is measured in joules (J). Energy is also measured in joules. Power is the ratio of total work done to the total time taken: Pavg = Δw/Δt.Power is measured in watts.Power is the time derivative of work: P = dW/dt. -4327283117850Useful Physics FormulasVelocity = d/dt position vector formulaPosition vector s = ∫ velocity dtAcceleration = d/dt velocity = d2/dt2 positionVelocity = ∫ acceleration dtJerk = d/dt acceleration = d2/dt2 velocity = d3/dt3 positionAcceleration = ∫ jerk dtUseful Physics FormulasVelocity = d/dt position vector formulaPosition vector s = ∫ velocity dtAcceleration = d/dt velocity = d2/dt2 positionVelocity = ∫ acceleration dtJerk = d/dt acceleration = d2/dt2 velocity = d3/dt3 positionAcceleration = ∫ jerk dt-427355-86360Derivatives of Transcendental Functionsd(sin(u))/dx = cos(u) dudxd(cos(u))/dx = –sin(u) dudxd(bu)/dx = bu ln(b) dudxd(tan(u))/dx = sec2(u) dudxd(cot(u))/dx = csc2(u) dudxd(eu)/dx = eu dudxd(sec(u))/dx = sec(u)tan(u) dudxd(csc(u))/dx = csc(u)cot(u) dudxd(logbu)/dx = 1/(u ln(b)) dudxd(sin–1(u))/dx = 1/1–u2 dudxd(cos–1(u))/dx = –1/1–u2 dudxd(ln(u))/dx = 1/u dudxd(tan–1(u))/dx = 1/1+u2 dudxd(cot–1(u))/dx = –1/1+u2 dudxd(sec–1(u))/dx = 1/|u|1–u2 dudxd(cos–1(u))/dx = –1/|u|1–u2 dudxd(sinh(u))/dx = cosh(u) dudx∫ sinh(u) du = cosh(u) + kd(cosh(u))/dx = sinh(u) dudx∫ cosh(u) du = sinh(u) + kd(tanh(u))/dx = 1/cosh2(u) dudx∫ tanh(u) du = sinh(u) / cosh(u) + kd(sinh-1 (u)/dx = 1/u2+1 dudx∫ sinh-1 (x) dx = xsinh-1 (x) – (x2+1) + kd(cosh-1 (u)/dx = 1/u2-1 dudx∫ cosh-1 (x) dx = xcosh-1 (x) – (x2-1) + kd(tanh-1 (u)/dx = 1/(1 – u2) dudx∫ tanh-1 (x) dx = ?ln(1 – x2) + xtanh-1 (x) + kDerivatives of Transcendental Functionsd(sin(u))/dx = cos(u) dudxd(cos(u))/dx = –sin(u) dudxd(bu)/dx = bu ln(b) dudxd(tan(u))/dx = sec2(u) dudxd(cot(u))/dx = csc2(u) dudxd(eu)/dx = eu dudxd(sec(u))/dx = sec(u)tan(u) dudxd(csc(u))/dx = csc(u)cot(u) dudxd(logbu)/dx = 1/(u ln(b)) dudxd(sin–1(u))/dx = 1/1–u2 dudxd(cos–1(u))/dx = –1/1–u2 dudxd(ln(u))/dx = 1/u dudxd(tan–1(u))/dx = 1/1+u2 dudxd(cot–1(u))/dx = –1/1+u2 dudxd(sec–1(u))/dx = 1/|u|1–u2 dudxd(cos–1(u))/dx = –1/|u|1–u2 dudxd(sinh(u))/dx = cosh(u) dudx∫ sinh(u) du = cosh(u) + kd(cosh(u))/dx = sinh(u) dudx∫ cosh(u) du = sinh(u) + kd(tanh(u))/dx = 1/cosh2(u) dudx∫ tanh(u) du = sinh(u) / cosh(u) + kd(sinh-1 (u)/dx = 1/u2+1 dudx∫ sinh-1 (x) dx = xsinh-1 (x) – (x2+1) + kd(cosh-1 (u)/dx = 1/u2-1 dudx∫ cosh-1 (x) dx = xcosh-1 (x) – (x2-1) + kd(tanh-1 (u)/dx = 1/(1 – u2) dudx∫ tanh-1 (x) dx = ?ln(1 – x2) + xtanh-1 (x) + k ................
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