Determine the equation of the tangent line to the curve defined by #(2x^4)(4y^4)+6x^3+7y^2=2703# at the point #(2,−3)#?

Question

815 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-14T23:48:59

The point #(2,-3)# does not lie on the given curve.

Put the coordinates #(2,-3)# into the given equation we get:

# LHS = 2(16)(4)(81)+6(8)+7(9) #
# \ \ \ \ \ \ \ \ = 10368 +48+63#
# \ \ \ \ \ \ \ \ = 10479#
# \ \ \ \ \ \ \ \ != 2703 #

So the point #(2,-3)# does not lie on the given curve.

Answer 2 · 2018-02-14T23:53:02

#y=-(3468x)/2311+3/2311#

#y=-1.5x-0.0013#

First off, we take #d/dx# of every term.

#d/dx[8x^4y^4]+d/dx[6x^3]+d/dx[7y^2]=d/dx[2703]#

#8y^4d/dx[x^4]+8x^4d/dx[y^4]+18x^2+d/dx[7y^2]=0#

#8y^4(4x^3)+8x^4d/dx[y^4]+18x^2+d/dx[7y^2]=0#

#32y^4x^3+8x^4d/dx[y^4]+18x^2+d/dx[7y^2]=0#

The chain rule gives us that:
#d/dx=dy/dx*d/dy#

#32y^4x^3+8x^4dy/dx d/dy[y^4]+18x^2+dy/dx d/dy[7y^2]=0#

#32y^3x^3+dy/dx[8x^4(4y^3)]+18x^2+dy/dx[14y]=0#

#dy/dx[32y^3x^4+14y]=-(18x^2+32y^4x^3)#

#dy/dx=-(18x^2+32y^4x^3)/(32y^3x^4+14y)#

Now we put in #x=2#, #y=-3#
#dy/dx=-(18(2)^2+32(-3)^4(2)^3)/(32(-3)^3(2)^4+14(-3))#
#color(white)(dy/dx)=-3468/2311# (will be converted later)

Equation of a tangent is #y=mx+c#

#-3=2(-3468/2311)+c#

#c=-3-2(-3468/2311)=3/2311#

#y=-(3468x)/2311+3/2311#

#y=-1.5x-0.0013#