MHB Solving for d^2y/dt^2: Finding d^2y/dt^2 with dx/dt=1/2

[riri]

Hello!

I'm confused on a question that states: by pythagorus, sin^2x+cos^2x=1 for any x. Suppose that y=-cos^4x=sin^4x.
If d^2x/dt^2=0, find d^2y/dt^2 when dx/dt=1/2

I first found dy/dt = 2sinxcosx
and ended up with $$\d{^2y}{t^2}$$ = cos^2x-sin^2x, and I was wondering if this is right?

 

[MarkFL]

You forgot about the factor of 2 when you obtained the second derivative...you should have gotten:

$$\d{^2y}{t^2}=2\left(\cos^2(x)-\sin^2(x)\right)$$

And then using a double-angle identity for cosine, you could write:

$$\d{^2y}{t^2}=2\cos(2x)$$

This is how I would work the problem:

We are given:

$$y=-\cos^4(x)+\sin^4(x)$$

Thus (differentiating with respect to $t$):

$$\d{y}{t}=-4\cos^3(x)\left(-\sin(x)\d{x}{t}\right)+4\sin^3(x)\left(\cos(x)\d{x}{t}\right)=4\sin(x)\cos(x)\d{x}{t}\left(\cos^2(x)+\sin^2(x)\right)=2\sin(2x)\d{x}{t}$$

And so, (differentiating again with respect to $t$):

$$\d{^2y}{t^2}=2\sin(2x)\d{^2x}{t^2}+4\cos(2x)\d{x}{t}$$

Now, we are told:

$$\d{^2x}{t^2}=0$$ and $$\d{x}{t}=\frac{1}{2}$$

And so we have:

$$\d{^2y}{t^2}=2\sin(2x)(0)+4\cos(2x)\left(\frac{1}{2}\right)=2\cos(2x)$$