Derivative of s with respect to t when y is constant

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Discussion Overview

The discussion revolves around the differentiation of the function \( s = \sqrt{6x^2 + 2y^2} \) with respect to time \( t \), specifically when \( y \) is treated as a constant. Participants explore the relationship between \( \frac{ds}{dt} \) and \( \frac{dx}{dt} \), applying the chain rule and addressing the implications of \( y \) being constant.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how \( \frac{ds}{dt} \) is related to \( \frac{dx}{dt} \) when \( y \) is constant.
  • Another participant suggests applying the chain rule to find \( \frac{ds}{dt} \).
  • A participant presents a formula for \( \frac{ds}{dt} \) but expresses uncertainty about the negative sign in front of the \( 6 \) and questions the implications of \( y \) being constant.
  • It is noted that if \( y \) is constant, then \( \frac{dy}{dt} = 0 \), leading to a simplification in the expression for \( \frac{ds}{dt} \).
  • Some participants discuss the simplification of the denominator and question whether \( 2y^2 \) equals zero, clarifying that while it does not equal zero, the term vanishes due to multiplication by \( \frac{dy}{dt} = 0 \).
  • There is a request for clarification on how the term \( 2y\frac{dy}{dt} \) vanishes to zero.

Areas of Agreement / Disagreement

Participants generally agree that \( \frac{dy}{dt} = 0 \) when \( y \) is constant, but there is no consensus on the implications for the overall expression for \( \frac{ds}{dt} \) or the simplification of the denominator.

Contextual Notes

Participants express uncertainty regarding the derivation steps and the treatment of constants in differentiation, particularly concerning the simplification of terms in the expression for \( \frac{ds}{dt} \).

karush
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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?
 
Last edited:
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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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