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

Answer 1

#slope =-1/0=-oo#

the given #sqrt(y*e^(x-y))=C# at #(-2,1)#

squaring both sides of the equation it becomes

#y*e^(x-y)=c^2#

differentiation both sides of the equation

#y'e^(x-y)+y*(e^(x-y))(1-y')=0#
#y'e^(x-y)+ye^(x-y)-ye^(x-y)*y'=0#
#y'(e^(x-y)-ye^(x-y))=-ye^(x-y)#
#y'=(-y(e^(x-y)))/((e^(x-y))(1-y))#
#y'=-y/(1-y)#
Evaluating #y'# at #y=1#
#y' = -1/(1-1)=-1/0=-oo#
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Answer 2

To find the slope of the tangent line to the curve defined by (\sqrt{ye^{x-y}} = C) at the point ((-2, 1)), we'll first need to differentiate both sides of the equation with respect to (x) using implicit differentiation. Given that (C) is a constant, its derivative will be (0). Let's solve this step by step.

The equation is given by: [ \sqrt{ye^{x-y}} = C ]

Squaring both sides to get rid of the square root yields: [ ye^{x-y} = C^2 ]

Differentiating both sides with respect to (x), using the product rule and chain rule where necessary, gives: [ \frac{d}{dx}\left(ye^{x-y}\right) = \frac{d}{dx}\left(C^2\right) ]

Since (C^2) is a constant, its derivative with respect to (x) is (0), so we have: [ \frac{d}{dx}\left(ye^{x-y}\right) = 0 ]

Let's perform the differentiation and solve for (\frac{dy}{dx}), which is the slope of the tangent line at the given point ((-2, 1)).The slope of the tangent line to the curve at any point is given by (\frac{y}{y - 1}). To find the slope at the specific point ((-2, 1)), we substitute (y = 1) into this expression. Let's calculate that.The calculation results in an undefined value, indicating that at the point ((-2, 1)), the slope of the tangent line to the curve defined by (\sqrt{ye^{x-y}} = C) is not defined due to a division by zero in the formula for the slope. This suggests that at this specific point, the behavior of the curve is such that a tangent line in the traditional sense does not exist or is vertical.

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Answer 3

To find the slope of the tangent line of the curve defined by the equation ( \sqrt{ye^{x-y}} = C ) at the point (-2,1), we need to first find the derivative dy/dx, then evaluate it at the given point.

Differentiating implicitly with respect to x, we get:

[ \frac{d}{dx} \left( \sqrt{ye^{x-y}} \right) = \frac{d}{dx} C ]

Using the chain rule and the fact that ( \frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot \frac{d}{dx} f(x) ), we have:

[ \frac{1}{2\sqrt{ye^{x-y}}} \cdot \left( y'e^{x-y} + ye^{x-y} - ye^{x-y} \right) = 0 ]

Simplifying, we find:

[ \frac{y'e^{x-y}}{2\sqrt{ye^{x-y}}} = 0 ]

[ y' = 0 ]

Now, to find the slope of the tangent line at (-2,1), we substitute x = -2 and y = 1 into the derivative we found:

[ y' = 0 ]

Therefore, the slope of the tangent line at the point (-2,1) is 0.

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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.

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.

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.

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.

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