Related Rates challenge problem, I think I understand it but I can't help but feel that the answer itself is slightly wrong. (One minute.)

Question

mendicant_bias · Answer

This is what the answer is supposed to be, which I think may be wrong:

anonymous · Answer

why?
i thind the develoment of the problem is right

mendicant_bias · Answer

This is what I did:

$$A = xy$$
$$\frac{ dA }{ dt } = y (\frac{ dy }{ dt }) + x (\frac{ dx }{ dt })$$
$$f(x) = x ^{3}-2xy+y ^{3}+1 = 0$$
$$f'(x) = 3x ^{2}(\frac{ dx }{ dt })  -2y (\frac{ dx }{ dt }) - 2x (\frac{ dy }{ dt })+3y ^{2}(\frac{ dy }{ dt }) = 0$$
One sec, coming back to this.

mendicant_bias · Answer

I think the development is correct, too, it's just at the very end that I take issue. Shouldn't, at the end, since we're ultimately looking for dA/dt, we plug back into that? After finding dy/dt, and given the x, y, and dx/dt values, shouldn't we plug back in? They find the value of dy/dt and then say that that is dA/dt.

anonymous · Answer

yes there is a step that they r missing

mendicant_bias · Answer

Made a mistake in rewriting out the last derivative, but that doesn't change my point. But yeah, it can't be that, right? It's hopefully just a writing error.

mendicant_bias · Answer

$$f'(x) = 3x ^{2}(\frac{ dx }{ dt }) -2y ^{2}(\frac{ dx }{ dt })-4xy(\frac{ dy }{ dt })+3y ^{2}(\frac{ dy }{ dt }) = 0$$
$$\frac{ dy }{ dt }=\frac{ (3x ^{2}-2y ^{2})(\frac{ dx }{ dt }) }{ 4xy-3y ^{2} }$$
Plugging in now for dy/dt = 1, x = 2, y = 3:

$$\frac{ dy }{ dt }=\frac{ [3(2)^{2}-2(3)^{2}](1) }{ 4(2)(3)-3(3)^{2} }$$
$$\frac{ dy }{ dt }=\frac{ -6 }{ -3 } = 2$$
Whoah, so, their dy/dt value is wrong, too? Jesus.

mendicant_bias · Answer

$$\frac{ dA }{ dt }= (1)(3)+(2)(2)=7$$Wait, so....the dA/dt value they specified is correct...it's just the dy/dt value that they fudged. Am I seeing this right?

phi · Answer

looks ok except you have a typo   and want dy/dt = 2

and you should put in one more line to show dA/dt = 7

amistre64 · Answer

Area of rectangle$$A=xy$$
$$A'=x'y+xy'$$they give you x', x, and y
$$A'=1(3)+2y'$$
use the controlling function (it controls the x and y values) to find y'
$$x^3-2xy^2+y^3+1=0$$
$$3x^2x'-2x'y^2-4xyy'+3y^2y'+0=0$$
x'=1, x=2, y=3
$$3(2)^2-2(3)^2-4(2)(3)y'+3(3)^2y'+0=0$$
solve for y'
$$12-18-24y'+27y'=0$$
$$12-18+3y'=0$$
$$4-6+y'=0$$
y' = 2

plugging that into the Area we get:$$A'=1(3)+2(2)=7$$

mendicant_bias · Answer

@phi: I did, I just put it in the next post, and I did write that dy/dt = 2.Thanks, though. guys.