Second Derivative of Parametric Equation

Calculus Level 3

Let $x,y : \mathbb{R} \to \mathbb{R}$ be functions satisfying $y(t) = \sin^2 t$ and $x(t) = \cos t$ for all $t$ . The second derivative $\frac{d^2 y}{d x^2}$ is a constant function, equal to $a$ everywhere. Find the value of $a$ .

The answer is -2.

1 solution

Jaime Cabrera
Feb 27, 2016

As we have to find $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ , first we get $x^2=\cos^2 t$

$y=\sin^2 t=1-\cos^2 t=1-x^2$ $y'=-2x$ $\boxed{y''=-2}$

I admit I can't put my finger on it, but there is something fishy about this answer. We are given parametric functions (presumably single-valued) for x and y. Y can only take on values of 0 to 1, while x can range from +1 to -1. This imposes boundaries on the possible shape of (x, y) points, a 1 unit by 2 unit rectangle. Outside this area, y'' is meaningless, or non-existent, not -2.

To elaborate further, this is posed as a parametric differentiation problem, yet there is none in the solution. So, while the answer may be correct, it was found using non-parametric differentiation. I tried to use parametric differentiation as follows:

y = (sin t)^2, x = cos t

dy/dt = 2 sin t cos t, dx/dt = -sin t

Then, dy/dx = -2 cos t, and d(dy/dx)/dx = 2 sin t dt/dx = 2 sin t/(-sin t) = -2

or, more simply, dy/dx = -2 x, y'' = -2

Tom Capizzi - 5 years, 1 month ago

Second Derivative of Parametric Equation

The answer is -2.

1 solution

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