Question

Suppose that a point moves along some unknown curve y = f(x) in such a way that at each point (x,y) on the curve the tangent line has a slope
`(x^(2))/(2)` . Find an equation for the curve,given that it passes through (1,1).

Answer 1

Solving for a Function given its derivative and a point

Answer:

`y = (x^3 +5)/(6)\\`

Explanation:

`(x^2)/(2)\\` is just the the derivative of
`f(x)` since it it is the slope of the line tangent to the curve at
`x`. To find how
`f(x)` is defined, we'll just take the anti-derivative of
`(x^2)/(2)\\`.

Recall:

`\int x^n \mathrm{d}x = (x^(n +1))/(n +1) &, &n \\eq -1\\`

Solving for
`f(x)`:

`f(x) = \int (x^2)/(2) \mathrm{d}x \\ f(x) = \frac12 \int x^2 \mathrm{d}x \\ f(x) = \frac12 \cdot (x^3)/(3) \\ f(x) = (x^3)/(6) +C`

We still have to solve for
`C`. Note that the point
`(1,1)` has to be on our graph so
`f(1) = 1` or
`((1)^3)/(6) +C = 1\\`.

Solving for
`C`:

`((1)^3)/(6) +C = 1 \\ (1)/(6) +C = 1 \\ C = 1 -(1)/(6) \\ C = (6)/(6) -(1)/(6) \\ C = (5)/(6)`

We can finally write
`f(x) = (x^3)/(6) +(5)/(6)\\` or
`f(x) = (x^3 +5)/(6)\\`. Since
`y = f(x)`, we can write it as
`y = (x^3 +5)/(6)\\` as well.