Question

What is the equation of the tangent line of `f(x) =x/sqrt(64-x^2)+(64-x^2)^(3/2)` at `x = 4`?

Answer 1

`0.19245x-y+332.3613=0`

Explanation:

Given:
`f(x)=x/sqrt(64-x^2)+(64-x^2)^(3/2)`
To find the equation of the tangent at `" x=4"`
Let `" "y=f(x)`
Now,`" "y=x/sqrt(64-x^2)+(64-x^2)^(3/2)`
Substituting for x=4,
`y=4/sqrt(64-4^2)+(64-4^2)^(3/2)`

`y=4/sqrt(64-16)+(64-16)^(3/2)`

`y=4/sqrt48+48^(3/2)`

The point through which the tangent passes is given by
`P-=(x,y)=P-=(4,333.1311)`

Differentiating wrt x

`dy/dx=((sqrt(64-x^2)-x/(2(sqrt(64-x^2)))(-2x))/(sqrt(64-x^2))^2)`
`+3/2(64-x^2)^(1/2)(-2x)`

`dy/dx=(64-x^2+x^2)/(64-x^2)^(3/2)`

`dy/dx=64/(64-x^2)^(3/2)`

Substituting x=4

`dy/dx=64/(64-4^2)^(3/2)`

`dy/dx=64/48^(3/2)`

`dy/dx=0.19245`

Slope of the tangent passing through the point `" "P-=(4,333.1311)` is 0.19245

Thus, the point slope form of the equation happens to be

`(y-333.1311)/(x-4)=0.19245`

Simplifying

`y-333.1311=0.19245(x-4)`

`y-333.1311=0.19245x-0.7698`

`y=0.19245x+332.3613`
or
`0.19245x-y+332.3613=0`