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

[Dethrone]

Since it's the summer, I might as well take advantage of all the math helpers on the site (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?

 

[MarkFL]

You method is solid, but I get a different value for \(\displaystyle \d{y}{t}\). Can you show your work?

 

[Dethrone]

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.

 

[MarkFL]

Okay, good, that's the value I found as well. :D

 

[Dethrone]

The rest is easy now:

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

How do I include text in $\LaTeX$?

 

[I like Serena]

The rest is easy now:

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

How do I include text in $\LaTeX$?

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

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

The dollars make sure that the square is in math mode again.

 

[MarkFL]

Here's another approach:

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

\(\displaystyle \d{A_r}{t}=7\frac{\text{units}^2}{\text{s}}\)