Question

How to get the values of a and b? i need to differentiate first the equation? ​

Answer 1

`a=-6`

`b=-(31)/(6)`

Explanation:

We are given the equation of the curve:

`y=2x^(3)+bx-5` (1)

And the equation of the line that is normal (perpendicular to that curve) in the point
`(x,y)=(1,2)`:

`y=(a)/(5)x+(16)/(5)` (2)

Note this equation has the form
`y=mx+B`, where
`m=(a)/(5)` is the slope of the line that es normal to the curve and
`B=(16)/(5)` the intersection of that line with the y-axis.

Well, let’s begin by finding the slope of the line that is tangent to the curve by differentiating the equation (1):

`(dy)/(dx)=6x^(2)+b` (3)

Then, we have to evaluate that when
`x=1` (remember we are given the point
`(1,2)`):

`(dy)/(dx)_(x=1)=6(1)^(2)+b=6+b` (4) This is the slope of the line that is tangent to the curve

And the slope of the line that is perpendicular to that tangent line is its negative reciprocal. Hence:

`m=-(1)/(6+b)` (5) This is the slope of the line normal to the curve

Then:

`m=(a)/(5)=-(1)/(6+b)`

`a=\-frac{5}{6+b}` (6)

Rewritting (2):

`y=-(5)/(5(6+b))x+(16)/(5)` (7)

Evaluating in the point
`(1,2)`:

`2=-(5)/(5(6+b))1+(16)/(5)` (8)

Finding
`b`:

`b=-(31)/(6)` (9) This is the value of
`b`

Substituting (9) in (6):

`a=\-frac{5}{6-(31)/(6)}` (10)

`a=\-frac{31}{6}` (11) This is the value of
`a`