Posted by **Aric** on Sunday, February 20, 2011 at 11:50am.

Find the slope of the tangent line to the curve (a lemniscate)

2(x^2+y^2)^2 = 25(x^2-y^2)

it is at pt (3,1)

- Calculus - Verify? -
**Aric**, Sunday, February 20, 2011 at 11:59am
:4(x² + y²)(2x + 2y dy/dx) = 25(2x - 2y dy/dx).

:4[(3)² + (1)²][2(3) + 2(1)dy/dx] = 25[2(3) - 2(1)dy/dx]

:4[9 + (1)][6 + 2 dy/dx] = 25[6 - 2 dy/dx]

:4[10][8 dy/dx] = 25[4 dy/dx]

Divide both sides by 10

:4[8 dy/dx] = 10[dy/dx]

:32[dy/dx] = 10 [dy/dx]

then I got dy/dx = 10/32

- Calculus - Verify? -
**Reiny**, Sunday, February 20, 2011 at 12:15pm
You messed up near the end

look at line

4[9 + (1)][6 + 2 dy/dx] = 25[6 - 2 dy/dx]

and then

4[10][8 dy/dx] = 25[4 dy/dx]

you added "unlike terms" inside the bracket

should have been

40[6+2dy/dx] = 25[6-2dy/dx]

240 + 80dy/dx = 150 - 50dy/dx

130dy/dx = -90

dy/dx = -90/130 = -9/13

the rest is easy.

- Calculus - Verify? -
**Aric**, Sunday, February 20, 2011 at 12:33pm
tyvm <3

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