Question

Consider the curve where a and b are constants. The normal to this curve at the point where x=4 is 4x+y=22. Find the values of a and b?

`y=asqrtx+ b/sqrtx`

Answer 1

The answer will be `a = 2`, `b = 4`

Explanation:

We can start by rewriting the function as

`y = a(x)^(1/2) + bx^(-1/2)`

Taking the derivative, and treating `a` and `b` as the constants, we get:

`y' = 1/2a(x)^(-1/2) - 1/2bx^(-3/2)`

Let's evaluate `x = 4` in the derivative to get the slope of the tangent line.

`y'(4) = 1/2(a)(4)^(-1/2) - 1/2(b)(4^(-3/2))`

`y'(4) = 1/2(a)(1/4)^(1/2) - 1/2(b)(1/4)^(3/2)`

`y'(4) = 1/4a - 1/16b`

That means that the slope of the normal line is the negative reciprocal or

`m_"normal" = -1/(1/4a - 1/16b)`

We're given the actual equation of the normal line is `4x + y = 22`. This means that `y = 22- 4x` and that the slope is `-4`. Therefore, our first equation will be

`4 = 1/(1/4a - 1/16b)`

At `x = 4`, given `y = 22 - 4x`, the normal lien will have value `y = 22- 16 = 6`.

Therefore:

`a(4)^(1/2) + b(4)^(-1/2) = 6`
`2a + 1/2b = 6`
`4a + b = 12`
`b = 12 - 4a`

Substituting this second equation into the first, we get:

`4 = 1/(1/4a - 1/16(12 - 4a))`

`4 = 1/(1/4a - 3/4 + 1/4a)`

`4 = 1/(1/2a - 3/4)`

`4(1/2a - 3/4) = 1`

`2a - 3 = 1`

`2a = 4`

`a = 2`

It follows that

`b = 12 - 4(2) = 4`

A graphical confirmation confirms that our answer is viable.

Hopefully this helps!