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What is the slope of the tangent line of # (x-2)(y-3)-e^y= C #, where C is an arbitrary constant, at #(-2,1)#?

Answer 1

Evelyn Agard

# y'|_{(-2,1)} = - (2)/(e +4)#

We can use implicit differentiation (and the product rule for the first term):

#(y-3) + (x-2) y' - y' e^y= 0#

#(y-3) = y' (e^y - x + 2)#

# y' = (y-3)/(e^y - x + 2)#

# y'|_{(-2,1)} = (1-3)/(e^1 - (-2) + 2) = - (2)/(e +4)#

we can also check this by evaluating the normal vector #mathbf n# as

#mathbf n = nabla ((x-2)(y-3)-e^y) = langle y-3, x-2 - e^y rangle#

#mathbf n_{(-2,1)} = langle -2, -4 - e rangle#

and for tangent vector #mathbf t#, we say:

#mathbf n * mathbf t = 0 implies \mathbf t = langle 4 + e, -2, rangle#, same answer

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Answer 2

Charles Arocha

To find the slope of the tangent line at (-2,1), first, differentiate the given equation implicitly with respect to x. Then, substitute the coordinates of (-2,1) into the derivative expression to find the slope.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.