Derivatives related

**bobsanchez** · July 25th 2010, 11:06 AM

If x^2 +y^2 =25 and dy/dt = 6, find dx/dt when y = 4

When I did this, I got positive and negative three, but that can't be right. What is the right way to do this?

**apcalculus** · July 25th 2010, 11:37 AM

The curve is the circle of radius 5, at y=4, which means x could be -3 or 3.

Fully differentiate:

2x x' + 2y y' = 0

x x' + y y' = 0

x x' + 4*6 = 0

x' = 24/x

so dx/dt is 8 or -8.

Only one answer is expected here?

**Archie Meade** · July 25th 2010, 12:44 PM

Originally Posted by bobsanchez

If x^2 +y^2 =25 and dy/dt = 6, find dx/dt when y = 4

When I did this, I got positive and negative three, but that can't be right. What is the right way to do this?

$x^2+y^2=5^2$
is a circle centred at the origin (0,0) with radius=5.

Hence, when y=4,

$x^2+16=25\ \Rightarrow\ x^2=9\ \Rightarrow\ x=\pm3$

The radius is unchanging..

$\frac{d}{dt}\left(x^2+y^2\right)=0$

$\frac{dx}{dt}\frac{d}{dx}x^2+\frac{dy}{dt}\frac{d} {dt}y^2=0$

$2x\frac{dx}{dt}=-2y\frac{dy}{dt}$

Solve this at (3,4) and (-3,4).

**bobsanchez** · July 27th 2010, 01:55 PM

No, two answers were expected.

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