MHB Derivative of s with respect to t when y is constant

[karush]

Let $x$ and $y$ be differentiable functions of $t$, and let
$$s=\sqrt{6x^2+2 y^2 }$$ be a function of $x$ and $y$.
A. How is ds/dt related to dx/dt if y is constant?

 

[MarkFL]

What would you get if you apply the chain rule to find $$\d{s}{t}$$?

 

[karush]

$$\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$$

I got ? on this

 

[MarkFL]

$$\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$$

I got ? on this

Where did the negative sign in front of the 6 come from?

If $y$ is a constant, then what is $$\d{y}{t}$$? What is the denominator equal to?

 

[karush]

A constant would go to zero

$\frac{ds}{dt}=\frac{6x\d{x}{t}+(0)}{\sqrt{6x^2 +2 y^2 }}$

 

[MarkFL]

Yes, we would have $$\d{y}{t}=0$$...how about that denominator? Can you replace it with something simpler?

 

[karush]

Does the $2y^2 = 0\ \ $ ?

 

[MarkFL]

Does the $2y^2 = 0\ \ $ ?

No, but since it is being multiplied by $$\d{y}{t}=0$$, the entire term does vanish.

 

[karush]

How does that happen?

 

[MarkFL]

How does that happen?

I'm sorry, I meant the term:

$$2y\d{y}{t}$$

in the numerator vanishes to zero. The denominator of your result can be simplified in another way, if you look at the problem statement, before you did any differentiating.