What is the rate of change for the area of the rectangle?

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Discussion Overview

The discussion revolves around the rate of change of the area of a rectangle whose upper right corner lies on a specific curve. Participants explore related rates in the context of calculus, specifically focusing on how to derive the rate of change of area as the rectangle's dimensions change over time.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a problem involving a rectangle with its upper right corner on the curve defined by the equation $x^3-2xy^2+y^3+1=0$ and seeks to find the rate of change of the area as the rectangle moves.
  • The area of the rectangle is expressed as $A_r(x,y) = xy$, and the participant differentiates this with respect to time to find $\d{A_r}{t} = y \d{x}{t} + x \d{y}{t}$.
  • Another participant questions the calculation of $\d{y}{t}$ and requests to see the work leading to that value.
  • A subsequent reply provides the differentiation of the curve with respect to time, leading to a calculation of $\d{y}{t}$, which is found to be $2$ after correcting an earlier mistake.
  • One participant confirms agreement with the value of $\d{y}{t}$ as $2$.
  • The final calculations for $\d{A_r}{t}$ are presented, yielding a result of $7$ units² per second.
  • There are additional inquiries about formatting text in LaTeX, with suggestions on how to include units in the mathematical expressions.

Areas of Agreement / Disagreement

Participants generally agree on the value of $\d{y}{t}$ being $2$ and the final rate of change of the area being $7$ units² per second. However, there is some variation in the approach to calculating $\d{y}{t}$, indicating that multiple methods may be considered.

Contextual Notes

Participants express uncertainty in the differentiation steps and the assumptions made during the calculations, particularly regarding the application of the chain rule and the handling of terms in the differentiation of the curve equation.

Dethrone
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Since it's the summer, I might as well take advantage of all the math helpers on the site :cool: (yay!)

I'm pretty rusted when it comes to related rates, so it'd be great if someone checked my work :D

Problem:
A rectangle has two sides along the positive coordinate axes and its upper right corner lies on the curve: $x^3-2xy^2+y^3+1=0$. How fast is the area of the rectangle changing as the point passes the position $(2,3)$ if it is moving at $\d{x}{t}=1$ units per second?

Progress:
$$A_r (x,y)=xy$$

Differentiating both sides w.r.t $t$:
$$\d{A_r}{t}=y\d{x}{t}+x\d{y}{t}$$

Plugging what we know:
$$\d{A_r}{t}=(1)(3)+(2)\d{y}{t}$$

We still need $\d{y}{t}$, which is where I'm not sure how to get. The only thing I can think of is to differentiate the given equation of the curve with respect to time, and then plug in $x$, $y$ and $\d{x}{t}$ which I get $\d{y}{t}=\frac{6}{51}$. Is that correct?
 
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You method is solid, but I get a different value for $$\d{y}{t}$$. Can you show your work?
 
No problem :D

$$x^3-2xy^2+y^3+1=0$$

Differentiating with respect to $t$:
$$3x^2\d{x}{t}-2\d{x}{t}y^2+2y\d{y}{t}(-2x)+3y^2\d{y}{t}=0$$
$$3(4)(1)-2(1)(9)+(6)\d{y}{t}(-4)+3(9)\d{y}{t}=0$$
$$-6-24\d{y}{t}+27\d{y}{t}=0$$
$$\d{y}{t}=2$$

Found a mistake, accidentally forget a negative side somewhere.
 
Okay, good, that's the value I found as well. :D
 
The rest is easy now:

$$\d{A_r}{t}=(1)(3)+(2)\d{y}{t}$$
$$\d{A_r}{t}=(1)(3)+(2)(2)$$
$$\d{A_r}{t}=7$$ units^2 per second

How do I include text in $\LaTeX$?
 
Rido12 said:
The rest is easy now:

$$\d{A_r}{t}=(1)(3)+(2)\d{y}{t}$$
$$\d{A_r}{t}=(1)(3)+(2)(2)$$
$$\d{A_r}{t}=7$$ units^2 per second

How do I include text in $\LaTeX$?

Just say that it is "text".
Like this: \text{ units\$^2\$ per second}. ;)

$$\d{A_r}{t}=7 \text{ units$^2$ per second}$$

The dollars make sure that the square is in math mode again.
 
Here's another approach:

Use frac{\text{units}^2}{\text{s}}...e.g.:

$$\d{A_r}{t}=7\frac{\text{units}^2}{\text{s}}$$
 

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