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

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When differentiating the function \( s = \sqrt{6x^2 + 2y^2} \) with respect to \( t \) while keeping \( y \) constant, the derivative simplifies to \( \frac{ds}{dt} = \frac{6x \frac{dx}{dt}}{\sqrt{6x^2 + 2y^2}} \). The term involving \( y \) vanishes because \( \frac{dy}{dt} = 0 \) when \( y \) is constant. The negative sign in the original expression is incorrect since it arises from misunderstanding the differentiation process. The denominator remains as \( \sqrt{6x^2 + 2y^2} \), which does not equal zero as long as \( x \) and \( y \) are not both zero. Thus, the correct relationship shows that \( \frac{ds}{dt} \) is directly proportional to \( \frac{dx}{dt} \) when \( y \) is constant.
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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?
 
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What would you get if you apply the chain rule to find $$\d{s}{t}$$?
 
$$\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$$

I got ? on this
 
karush said:
$$\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$$

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?
 
A constant would go to zero

$\frac{ds}{dt}=\frac{6x\d{x}{t}+(0)}{\sqrt{6x^2 +2 y^2 }}$
 
Last edited:
Yes, we would have $$\d{y}{t}=0$$...how about that denominator? Can you replace it with something simpler?
 
Does the $2y^2 = 0\ \ $ ?
 
karush said:
Does the $2y^2 = 0\ \ $ ?

No, but since it is being multiplied by $$\d{y}{t}=0$$, the entire term does vanish.
 
How does that happen?
 
  • #10
karush said:
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.
 

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