Question

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

Answer 1

We have the following curve:


`y^2x^3-15x^2=4`

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for
`y`, thus, let's apply implicit differentiation, so:


`(d)/(dx)(y^2x^3-15x^2=4) \\ \\ (d)/(dx)(y^2x^3)-(d)/(dx)(15x^2)=(d)/(dx)(4) \\ \\ (y^(2))'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\ y'=(dy)/(dx)=(30x-3x^2y^2)/(2x^3y)=(x(30-3xy^2))/(2x^3y) \\ \\ y'=(30-3xy^2)/(2x^2y)`

Therefore the horizontal lines occurs when
`y'=0`, then:


`(30-3xy^2)/(2x^2y)=0 \\ \\ That \ is, \ when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=(10)/(x)`

If we substitute this in the original equation we have:


`(10)/(x)x^3-15x^2=4 \\ \\ \therefore 10x^2-15x^2=4 \\ \\ \therefore x^2=-(4)/(5)`

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of
`x` in which the curve has horizontal tangent lines.