Find the coordinates of the point on the parabola y = x^2 closest to the point (0, 4.5).
Answer and explanation
Correct answer: B. (2, 4) and (-2, 4)
- Step 1: Step 1: Distance squared S(x) = (x - 0)^2 + (y - 4.5)^2 = x^2 + (x^2 - 4.5)^2 = x^2 + x^4 - 9x^2 + 20.25 = x^4 - 8x^2 + 20.25. Step 2. Step 2: Differentiate: dS/dx = 4x^3 - 16x = 4x(x^2 - 4) = 0 => x = 0 or x = +/- 2. Second derivative: d^2S/dx^2 = 12x^2 - 16. At x = 0, S'' = -16 < 0 (local max). At x = +/- 2, S'' = 48 - 16 = 32 > 0 (minima). Points are (+/- 2, 4). Answer: (2, 4) and (-2, 4). Why not the others: Minimising squared distance avoids complex radical derivatives.
Practise more Differential and Integral Calculus, Real-World Modelling and Differential Equations.
Free questions with an explanation for every one, and full timed New Zealand Scholarship mock tests.
More Differential and Integral Calculus, Real-World Modelling and Differential Equations questions
- Evaluate the indefinite integral integral( 2x * cos(x^2) dx ) using substitution.
- Using implicit differentiation, find dy/dx for the curve x^2 + y^2 = 25 at the point (3, 4).
- What is the general solution of the separable differential equation dy/dx = 3x^2 * y for y > 0?
- Using integration by parts, evaluate integral( x * e^x dx ).
- What is the volume of the solid generated when the region under y = sqrt(x) from x = 0 to x = 4 is revolved 360 degrees about the x-axis?
- Differentiate y = ln(x^2 + 1) with respect to x.