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

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The discussion revolves around calculating the rate of change of the area of a rectangle with its upper right corner on the curve defined by the equation \(x^3-2xy^2+y^3+1=0\). The area \(A_r\) is expressed as \(xy\), and the differentiation with respect to time \(t\) leads to the equation \(\d{A_r}{t}=y\d{x}{t}+x\d{y}{t}\). After determining \(\d{y}{t}\) through implicit differentiation of the curve equation, the correct value is found to be \(2\). Substituting this back into the area equation results in a rate of change of the area \(\d{A_r}{t}=7\) units² per second. The discussion also touches on formatting text in LaTeX for clarity in mathematical expressions.
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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