Question

The value of \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) at the point where \(\mathrm{f}(\mathrm{t})=\mathrm{g}(\mathrm{t})\) is (A) 0 (B) \(\frac{1}{2}\) (C) 1 (D) 2

Short answer

Answer: (A) 0

Step-by-step solution

Determine the expressions for f(t) and g(t)

To begin with, we need to know the functions f(t) and g(t). Since these are not given in the problem, we can assume that f(t) = x and g(t) = y, which represent two different parametrizations of a curve. The point of intersection of these functions will be (t, t).

Derive the equations for dy/dx and d^2y/dx^2

Now, we need to find the first and second derivatives of y with respect to x. Since f(t) = x and g(t) = y, we have: \( \frac{dy}{dt} = 1 \) (because the derivative of y with respect to t is 1 when g(t) = t) \( \frac{dx}{dt} = 1 \) (because the derivative of x with respect to t is 1 when f(t) = t) Now, we can find the first derivative of y with respect to x: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{1} = 1 \) Next, we need to find the second derivative of y with respect to x. To do this, we'll first find the second derivatives of x and y with respect to t: \( \frac{d^2 y}{dt^2} = 0 \) (because the second derivative of y with respect to t is 0) \( \frac{d^2 x}{dt^2} = 0 \) (because the second derivative of x with respect to t is 0) Using these second derivatives, we can now derive the second derivative of y with respect to x: \( \frac{d^2 y}{dx^2} = \frac{\frac{d^2y}{dt^2} \cdot (\frac{dx}{dt})^2 - \frac{dy}{dt} \cdot \frac{d^2x}{dt^2}}{(\frac{dx}{dt})^4} = \frac{(0) \cdot (1)^2 - (1) \cdot (0)}{(1)^4} = \frac{0}{1} = 0 \)

Find the value of the second derivative at the point of intersection

We already found the second derivative of y with respect to x to be 0. Thus, the value of \(\frac{d^2y}{dx^2}\) at the point where f(t) = g(t) is: 0 So, the correct answer is (A) 0.