another problem

[warsatan]

not trying to flood the room tonight, but i got another problem, thanks again .

If the line 4x-9y=0 is tangent in the first quadrant to the graph of y = (1/3)x^3 + c , what's the value of c?

 

[Guest]

line 4x-9y=0 is tangent in the first quadrant to the graph of
y = (1/3)x^3 + c

4x- 9y =0

9y = 4x
y = (4/9) x
ie the gradient is (4/9)

y = (1/3)x^3 + c

y'(x) = x^2
This function y'(x) will have the same slope value as above (4/9)
then
(4/9) = x^2
(2/3) = x

Here is your x point where the tangent occurs
Put this value into the linear eqn to find the y value, then use this set of points in the fuction to find c
all the best

 

[warsatan]

ok, fully understand the problem now, thanks again.

ps , didnt know what 'gradient' mean, had to look it up , (slope)

 

[Guest]

sorry to use that term.. depends on where and when you did maths as to what you use.

gradient
slope
rate of change
tan theta
dy/dx
m
(y2-y1)/(x2-x1)

all are slope..... :shock: