Question

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

Answer 1

`c=16/81`

Explanation:

We'll have to do some thinking in order to solve this problem.

First of all, from a calculus point of view, what does "tangent" mean? Well, it's a line that touches a curve at one point. And what do we know about lines? They have constant slopes! But the question is asking us about `y=1/3x^3+c`, which is not a line - it's a curve. To find the slope of a curve, we need to take it's derivative, so we'll start there:
`y=1/3x^3+c`
`dy/dx=3*1/3x^(3-1)=x^2->` using the power rule

Ok, now where does that leave us? Well, we have an expression for the slope of the tangent line at a point (remember, that's what the derivative is). Wait a minute! Didn't the question tell us that the line `4x-9y = 0` is tangent to `y=1/3x^3+c`? In other words, the slope of the tangent line of `y=1/3x^3+c` is equal to the slope of `4x-9y=0` at the point where they are tangent (because `4x-9y=0` is the slope). If we add `9y` to both sides and divide by `9`, we see that `y=4/9x`. Let's set `4/9`, which is the slope, equal to `dy/dx` and see what happens:
`4/9=x^2->x=2/3`

Note that we take the positive square root because the point in question is in the first quadrant, as specified by the question.

This means that `4x-9y=0` is tangent to `y=1/3x^3+c` at the point `x=2/3`. To find the `y`-value of this point, we plug in to `y=4/9x`:
`y=4/9x=4/9*(2/3)=8/27`

So now we have a point, `(2/3, 8/27)`, and an unknown in `y=1/3x^3+c`. We can use our `x` and `y` values here to solve for `c`:
`y=1/3x^3+c`
`8/27=1/3(2/3)^3+c`
`8/27=8/81+c`
`c=8/27-8/81=16/81`