# Thread: "Decreasing" rate means - sign?

1. ## "Decreasing" rate means - sign?

Hi again all,

I answered a related rates type problem from a Calc final and the solution sheet's answer was +6 where my answer was -6:

The equation of a curve in the plane is e^(xy) = x^2 +y^2

A particle is moving along this curve at a certain velocity. At the instant that
it moves through the point (0, 1) the y coordinate is decreasing
at the rate of 3 cm/sec. How fast is the x coordinate changing
at this instant and is it increasing or decreasing.

I interpreted dy/dt as -3 cm/sec but in their answer sheet they marked it as 3, so would I get marks off for something like this?

2. Originally Posted by DannyMath
Hi again all,

I answered a related rates type problem from a Calc final and the solution sheet's answer was +6 where my answer was -6:

The equation of a curve in the plane is e^(xy) = x^2 +y^2

A particle is moving along this curve at a certain velocity. At the instant that
it moves through the point (0, 1) the y coordinate is decreasing
at the rate of 3 cm/sec. How fast is the x coordinate changing
at this instant and is it increasing or decreasing.

I interpreted dy/dt as -3 cm/sec but in their answer sheet they marked it as 3, so would I get marks off for something like this?
The answer they seek is the x coordinate is decreasing at 6 cm/s. You would probably get full marks if you sais the x-coordinate is changing at -6 cm/s, and that in consequence it is decreasing. Full or partial credit for other combinations of signs and words. (assuming that the marker is neither Attila the Hun nor Nogbad the bad).

CB

3. Originally Posted by CaptainBlack
The answer they seek is decreasing at 3 cm/s. You would probably get full marks if you sais the x-cordinate is changing at -3 cm/s, and that in consequence it is decreasing. Full or partial credit for other combinations of signs and words. (assuming that the marker is neither Attila the Hun nor Nogbad the bad).

CB
Yeah that's what I was thinking. And no worries! Marker's name is Adolf Stalin, seems like a nice chap!

4. Sorry, but the correct answer, using y'= -3, is 6, not -6.

You are given that $\displaystyle e^{xy}= x^2+ y^2$ so $\displaystyle ye^{xy}x'+ xe^{xy}y'= 2xx'+ 2yy'$. Setting x= 0, y= 1, y'= -3, that gives $\displaystyle 1(1)x'+ 0(1)(-3)= 2(1)x'+ 2(1)(-3)$
Then $\displaystyle x'- 2x'= -x'= -6$ so $\displaystyle x'= 6$.

5. Originally Posted by HallsofIvy
Sorry, but the correct answer, using y'= -3, is 6, not -6.

You are given that $\displaystyle e^{xy}= x^2+ y^2$ so $\displaystyle ye^{xy}x'+ xe^{xy}y'= 2xx'+ 2yy'$. Setting x= 0, y= 1, y'= -3, that gives $\displaystyle 1(1)x'+ 0(1)(-3)= 2(1)x'+ 2(1)(-3)$
Then $\displaystyle x'- 2x'= -x'= -6$ so $\displaystyle x'= 6$.
I think you might have substituted an x = 1 rather than x = 0 on the right hand side? I did it again and got -6, I was just confused cause on the answer sheet theylabeled dy/dt as 3 rather than -3 but I'm sure as long as I explain it on the exam I will be fine. Thanks though