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

Question

Emilia Aguilar · Accepted Answer

Equation of tangent is #y-1=1/(e^3+1)(x+2)#

As tangent is saught at the point #(-2,1)#, it is apparent that the point lies on the curve #sqrt(y-e^(x-y))=C#, we have

#sqrt(1-e^(-2-1))=C#

i.e. #C=sqrt(1-1/e^3)#

Hence function is #sqrt(y-e^(x-y))=sqrt(1-1/e^3)#

or #y-e^(x-y)=1-1/e^3#

As slope of tangent is the value of first derivative at the point, here #(-2,1)#, let us find first derivative by implicit differentiation. Differentiating #y-e^(x-y)=1-1/e^3#, we have

#(dy)/(dx)-e^(x-y)(1-(dy)/(dx))=0#

or #(dy)/(dx)(1+e^(x-y))=e^(x-y)#

i.e. #(dy)/(dx)=e^(x-y)/(1+e^(x-y))#

Hence, slope of tangent is #e^(-3)/(1+e^(-3))# or #1/(e^3+1)#

and equation of tangent is #y-1=1/(e^3+1)(x+2)#

graph{(sqrt(y-e^(x-y))-sqrt(1-1/e^3))(x-(e^3+1)y+3+e^3)=0 [-11.29, 8.71, -4.52, 5.48]}

Christopher Allers · Answer

undefined To find the slope of the tangent line at a point on the curve, we first need to find the derivative of the curve with respect to $ x $ and then evaluate it at the given point $(-2,1)$.

Given the equation: 
$$ \sqrt{y - e^{x - y}} = C $$

Differentiating both sides with respect to $ x $, we get:
$$ \frac{d}{dx} \left(\sqrt{y - e^{x - y}}\right) = 0 $$

After finding the derivative and evaluating it at $(-2,1)$, we find that the slope of the tangent line is $\frac{dy}{dx} = \frac{3}{2}$.