Topic: Engineering Mathematics
Equation of the line normal to function f(x) = (x - 8)^(2/3) + 1 at P(0,5) is
- A. y = 3x - 5
- B. 3y = x - 15
- C. 3y = x + 15
- D. y = 3x + 5
Answer: D. y = 3x + 5
Explanation: The slope of the tangent to the curve is found by taking the first derivative: m1 = dy/dx = (2/3) * (x - 8)^(-1/3). Evaluating this at x = 0 yields m1 = (2/3) * (-8)^(-1/3) = (2/3) * (-1/2) = -1/3. The slope of the normal line (m2) is the negative reciprocal of the tangent slope, so m2 = -1 / (-1/3) = 3. Using the point-slope form y = mx + c with the point (0,5), we substitute to get 5 = 3(0) + c, resulting in c = 5. The final equation is y = 3x + 5.