Find the equation(s) of the line(s) tangent to the curve y=x^(2/3) that are parallel to y=(1x/3)-5. Show calculus and algebra steps.

Question

Show all work. Thanks.

Answer 1

dy/dx=y'=(2/3)x^(2/3-1)=(2/3)x^(-1/3) or 2/3x^(1/3).

The slope of y=(x/3)-5 is 1/3 and this is the slope of all lines parallel to it.

The general equation of such lines is: y=(x/3)+a where a is a constant. 

So at a particular point on the curve 1/3=2/(3x^(1/3)), 1=2/x^(1/3), x^(1/3)=2 and x=2^3=8, making y=8^(2/3)=4.

The tangent line is given by y=x/3+a, where a is found by substituting for x and y:

4=8/3+a, so a=4-8/3=4/3 and y=(1/3)(x+4) is the tangent line.