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

Question

1295 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-12T09:50:03

slope #= 2#

Start with the given # (y/x)e^(x/y) = C# at #(1, 2)#

obtain the derivative of both sides of the equation

#(y/x)e^(x/y)*((y*1-x*y')/y^2)+e^(x/y)*((xy'-y*1)/x^2)=0#

After dividing both sides by #e^(x/y)#

#(y/x)*((y*1-x*y')/y^2)+((xy'-y*1)/x^2)=0#

#-y'/y+y'/x=y/x^2-1/x#

#y'(1/x-1/y)=y/x^2-1/x#

#y' = (y/x^2-1/x)/(1/x-1/y)#

substitute now the values #x=1# and #y=2#

#y' = (2/1^2-1/1)/(1/1-1/2)=(2-1)/(1-1/2)=2#