curve and line intersect at one point
x = 3
y = ?
substitute value for x into curve equation
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
Therefore, k = -27 ------answer.
umm....noo..wad im trying to say is that i am trying to construct the eqn of the tangent to the curve so as to compare with the eqn they gave us to find the value of k. But the problem is that i dunno how to simplify to get 15x-16y=k.( because my eqn has got sqr; very complicated=( )