solution:
y^2=x^3(26-x) @ (1,5)
2ydy=3x^2dx(26-x)+(-dx)x^3
[2ydy=3x^2dx(26-x)+(-dx)x^3]1/dx
2ydy/dx=3x^2(26-x)-x^3
dy/dx=[3x^2(26-x)-x^3]/2y
substitute (1,5)
dy/dx=[3(1)^2(26-1)-(1)^3]/2(5)
using calculator:
dy/dx=74/10 or 37/5
then dy/dx=m or slope: we have
m=37/5
equation of line is:
y-y1=m(x-x1) where y1 & x1 is y sub 1
substitute m=37/5 (1,5) to x1 & y1;we have
y-1=37/5(x-5)
y-1=37/5 x-37/25
y=37/5 x-37/25+1
y=37/5 x-12/25
[y=37/5 x-12/25](25)
25y=185x-12
185x-25y=12 (ans)
Use implicit differentiation to find an equation
of the tangent line to the curve
y^2 = x^3 (26 − x)
at the point (1, 5).
1 answer
1 answer