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 $\frac{x^{2}}{2}$ . 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:

$y = \frac{x^3 +5}{6}$

Step-by-step explanation:

$\frac{x^2}{2}$ 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 $\frac{x^2}{2}$.

Recall:

$\int x^n \mathrm{d}x = \frac{x^{n +1}}{n +1} &, &n \neq -1$

Solving for $f(x)$:

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

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

Solving for $C$:

$\frac{(1)^3}{6} +C = 1 \\ \frac{1}{6} +C = 1 \\ C = 1 -\frac{1}{6} \\ C = \frac{6}{6} -\frac{1}{6} \\ C = \frac{5}{6}$

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