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.
ScanSolve AI Answer
1.1 The sides x and y of a right-angled triangle are 2 mm and 3 mm respectively with the hypotenuse as 'r'. Calculate the approximate change in length of the hypotenuse if x changes to 3 mm and y changes to 5 mm. Step 1: Define the hypotenuse r and the initial conditions. The hypotenuse r of a right-angled triangle is given by the Pythagorean theorem: r = sqrt(x^2 + y^2) The initial values are x_0 = 2 mm and y_0 = 3 mm. The changes in x and y are: x = 3 mm - 2 mm = 1 mm y = 5 mm - 3 mm = 2 mm Step 2: Find the partial derivatives of r with respect to x and y. ( r)/( x) = ()/( x)(x^2 + y^2)^1/2 = (1)/(2)(x^2 + y^2)^-1/2(2x) = (x)/(sqrt(x^2 + y^2)) ( r)/( y) = ()/( y)(x^2 + y^2)^1/2 = (1)/(2)(x^2 + y^2)^-1/2(2y) = (y)/(sqrt(x^2 + y^2)) Step 3: Evaluate the partial derivatives at the initial values (x_0, y_0) = (2, 3). At (2, 3), the initial hypotenuse length is sqrt(2^2 + 3^2) = sqrt(4+9) = sqrt(13). ( r)/( x) |_(2,3) = (2)/(sqrt(13)) ( r)/( y) |_(2,3) = (3)/(sqrt(13)) Step 4: Calculate the approximate change in r using the total differential formula, dr = ( r)/( x) x + ( r)/( y) y. dr = ((2)/(sqrt(13)))(1) + ((3)/(sqrt(13)))(2) dr = (2)/(sqrt(13)) + (6)/(sqrt(13)) = (8)/(sqrt(13)) To rationalize the denominator: dr = (8)/(sqrt(13)) · sqrt(13)sqrt(13) = 8sqrt(13)13 The approximate change in the length of the hypotenuse is: 8sqrt(13)13 mm ≈ 2.22 mm 1.2 Given that x = t - t^2 and y = (3t), determine: 1.2.1 (dy)/(dx) Step 1: Find the derivatives of x and y with respect to t. (dx)/(dt) = (d)/(dt)(t - t^2) = 1 - 2t (dy)/(dt) = (d)/(dt)((3t)) = (1)/(3t) · 3 = (1)/(t) Step 2: Use the chain rule for parametric differentiation, (dy)/(dx) = (dy/dt)/(dx/dt). (dy)/(dx) = (1)/(t)1 - 2t (dy)/(dx) = (1)/(t(1 - 2t)) 1.2.2 (d^2 y)/(dx^2) Step 1: Use the formula for the second derivative of parametric equations: (d^2 y)/(dx^2) = (d)/(dt)((dy)/(dx)) · (dt)/(dx). From 1.2.1, we have (dy)/(dx) = (1)/(t(1 - 2t)) = (t - 2t^2)^-1. Also, (dt)/(dx) = (1)/(dx/dt) = (1)/(1 - 2t). Step 2: Calculate (d)/(dt)((dy)/(dx)). (d)/(dt)((t - 2t^2)^-1) = -1(t - 2t^2)^-2 · (d)/(dt)(t - 2t^2) = -(t - 2t^2)^-2 (1 - 4t) = -(1 - 4t)/((t(1 - 2t))^2) Step 3: Multiply the results from Step 1 and Step 2. (d^2 y)/(dx^2) = (-(1 - 4t)/((t(1 - 2t))^2)) · ((1)/(1 - 2t)) (d^2 y)/(dx^2) = -(1 - 4t)/(t^2(1 - 2t)^3) Last free one today — make it count tomorrow, or type /upgrade for unlimited.