Thread: finding unknown in eqn of the tangent to the curve

1. finding unknown in eqn of the tangent to the curve

The equation of a curve is y= 3x/sqr(1+x) . Given that the equation of the tangent to the curve at the point x=3 is 15x-16y=k, find the value of k.

I know roughly wad to do but after finding dy/dx which is the gradient of the curve as well as the tangent to the curve at the point x=3,
i had problems constructing the equation of the curve(it is too complicated, i dunno how to simplify)....

2. curve and line intersect at one point

x = 3
y = ?

substitute value for x into curve equation

3. Originally Posted by maybeline9216
The equation of a curve is y= 3x/sqr(1+x) . Given that the equation of the tangent to the curve at the point x=3 is 15x-16y=k, find the value of k.

I know roughly wad to do but after finding dy/dx which is the gradient of the curve as well as the tangent to the curve at the point x=3,
i had problems constructing the equation of the curve(it is too complicated, i dunno how to simplify)....
So you found that the slope of the tangent line to the curve at x=3 is 15/16?

And you can find the the equation of the tangent line at x=3?
Why, then you should be able to get the k.

Anyway, at x = 3,
y = (3*3) /sqrt(1+3) = 9/2
So, point (3,9/2)

The equation of the tangent line is
(y -9/2) = (15/16)(x -3)
Clear the fractions, multiply both sides by 16,
16y -72 = 15x -45
-72 +45 = 15x -16y
-27 = 15x -16y