This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here's the solution for question 7. 7. (a) A function y = 2x^4 - Px^3 + 2x, where P is a constant, has a stationary value at x = -1. Find the value of P. Step 1: Find the first derivative of the function with respect to x. For a stationary value, the derivative (dy)/(dx) must be equal to zero. y = 2x^4 - Px^3 + 2x (dy)/(dx) = (d)/(dx)(2x^4) - (d)/(dx)(Px^3) + (d)/(dx)(2x) (dy)/(dx) = 2(4x^4-1) - P(3x^3-1) + 2(1) (dy)/(dx) = 8x^3 - 3Px^2 + 2 Step 2: Set the derivative to zero and substitute x = -1. Since there is a stationary value at x = -1, we have (dy)/(dx) = 0 when x = -1. 0 = 8(-1)^3 - 3P(-1)^2 + 2 0 = 8(-1) - 3P(1) + 2 0 = -8 - 3P + 2 Step 3: Solve for P. 0 = -6 - 3P 3P = -6 P = (-6)/(3) P = -2 7. (b) The motion of a particle along a straight line is defined by the equation x = 5t^3 - 4t^2. Find after 4 seconds, the: (i) Velocity Step 1: Find the velocity function by differentiating the displacement x with respect to time t. x = 5t^3 - 4t^2 v = (dx)/(dt) = (d)/(dt)(5t^3) - (d)/(dt)(4t^2) v = 5(3t^3-1) - 4(2t^2-1) v = 15t^2 - 8t Step 2: Substitute t = 4 seconds into the velocity function. v = 15(4)^2 - 8(4) v = 15(16) - 32 v = 240 - 32 v = 208 units/s (ii) Acceleration Step 3: Find the acceleration function by differentiating the velocity v with respect to time t. v = 15t^2 - 8t a = (dv)/(dt) = (d)/(dt)(15t^2) - (d)/(dt)(8t) a = 15(2t^2-1) - 8(1) a = 30t - 8 Step 4: Substitute t = 4 seconds into the acceleration function. a = 30(4) - 8 a = 120 - 8 a = 112 units/s^2 7. (c) Evaluate _0^3 6x^2 - 4x + 2 \,dx. Step 1: Find the indefinite integral of the function. (6x^2 - 4x + 2) \,dx = 6x^2+12+1 - 4x^1+11+1 + 2x^0+10+1 + C = (6x^3)/(3) - (4x^2)/(2) + 2x + C = 2x^3 - 2x^2 + 2x + C Step 2: Evaluate the definite integral using the limits of integration from 0 to 3. [2x^3 - 2x^2 + 2x]_0^3 = (2(3)^3 - 2(3)^2 + 2(3)) - (2(0)^3 - 2(0)^2 + 2(0)) = (2(27) - 2(9) + 6) - (0 - 0 + 0) = (54 - 18 + 6) - 0 = 36 + 6 = 42 That's 2 down. 3 left today — send the next one.