posted by krishna on .
if the line 3x-4y=0 is tangent in the first quadrant to the curve y=x^3+k, find k.
all i know is that its related to finding a derivative and i m basically completely lost....can please someone provide me some help?????
the derivative gives the slope of the tangent line to the curve, at any point (x,y)
The line 3x-4y=0 has slope= 3/4
At any point on the curve (x,x^3 + k) the slope of the tangent line is 3x^2
So, you want to find x where 3x^2 = 3/4
x = 1/2
so, now you have y = 1/2^3 + k = 1/8 + k
What point on the line has y = 1/8 + k?
3* 1/2 - 4(1/8 + k) = 0
3/2 - 1/2 - 4k = 0
4k = 1
k = 1/4
So, y = x^3 + 1/4
At x = 1/2, y = 1/8 + 1/4 = 3/8
Note that at the point (1/2 , 3/8) 3x-4y = 0
Note on missing answer.
x can also be -1/2 if 3x^2 = 3/4
In that case, k = -1/4
and the point of tangency is (-1/2 , -3/8)
Note also, that if the line had been 3x-4y=1, the same line would have touched the curve at both points, with k=0.
thanks so much!