Final Answer
d^2y/dx^2 at the given point when x = sqrt(t), y = 2t + 4, t = 1 is 0.
Step-by-step explanation:
To find d^2y/dx^2 without eliminating the parameter, we'll first express y in terms of t and then differentiate twice with respect to x using the given relationship x = sqrt(t) and y = 2t + 4.
Given y = 2t + 4, to relate it to x, we're given x = sqrt(t). We can solve for t in terms of x: t = x^2. Then, substitute this expression for t into the equation for y, y = 2t + 4 = 2(x^2) + 4.
Now, to find d^2y/dx^2, differentiate y = 2(x^2) + 4 with respect to x. The first derivative dy/dx = d/dx(2x^2 + 4) = 4x. Taking the second derivative, d^2y/dx^2 = d/dx(4x) = 4.
Finally, evaluating d^2y/dx^2 at x = sqrt(t) when t = 1, substitute t = 1 into x = sqrt(t) to get x = 1. Therefore, d^2y/dx^2 = 4 at x = 1. Hence, at x = 1 (or t = 1), the second derivative d^2y/dx^2 = 4, indicating that the curve is a constant, and its curvature doesn't change at that point.