Calculus/Derivative

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Expert: - 1/9/2010


Question
Find the location of  all horizontal and vertical tangents...
1)). x * y^2 - 2y = 2      2)). x^2 + y^3 - 3y = 4

Please illustrate your answer.....

Answer
1) The equation can be solved for x.

It works out to b e xy² = 2y + 2, which then goes to x = 2(y+1)/y².
The derivative is then dx/dy = (2y² - 4y)/y^4.
This factors into 2(y-2)/y^3, with y=0 a point to look at, since we cancelled a y/y.

Where y=2, y-2=0, so dx/dy = 0, and since the derivative was with x with respect to y,
that means that this is a vertical tangent at y=2.  Put y=2 in the equation to find x.

If we set y = 0, we get that x is undefined, so there are no horizontal tangents.

2) The equation is x² = -y³ +3y + 4.
From here we can say that 2x dx = (-3y² + 3)dy,
so dy/dx = 2x/(-3y²+3) or dx/dy = -3(y²-1)/(2x).

Setting dy/dx=0 gives x=0, and that is where there is a vertical tangent.
Put x = 0 into the original equation and solve for .

dx/dy = 0 where y= ±1, and this can be put into the equation to find x.

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