If u is a function of v such that u is a constant plus a term that varies directly as the square of v. If u = 35 when v = 2, and u = 203 when v = 5, find v when u = 515.

The problem states that $u$ is the sum of two terms: one is a constant, and the other varies directly as the square of $v$. We can write this relationship as:

$$u = A + Bv^2$$

where $A$ is the constant term and $B$ is the constant of proportionality.

When $v = 2$, $u = 35$. Substitute these values into the equation:

$$35 = A + B(2)^2$$ $$35 = A + 4B \quad (*)$$

When $v = 5$, $u = 203$. Substitute these values into the equation:

$$203 = A + B(5)^2$$ $$203 = A + 25B \quad (**)$$

Subtract equation $(*)$ from equation $(**)$:

$$(A + 25B) - (A + 4B) = 203 - 35$$ $$21B = 168$$

Divide by 21:

$$B = \frac{168}{21}$$ $$B = 8$$

Substitute $B = 8$ into equation $(*)$:

$$35 = A + 4(8)$$ $$35 = A + 32$$ $$A = 35 - 32$$ $$A = 3$$

So, the relationship between $u$ and $v$ is $u = 3 + 8v^2$.

Substitute $u = 515$ into the derived equation:

$$515 = 3 + 8v^2$$

Subtract 3 from both sides:

$$515 - 3 = 8v^2$$ $$512 = 8v^2$$

Divide by 8:

$$v^2 = \frac{512}{8}$$ $$v^2 = 64$$

Take the square root of both sides. Since $v$ typically represents a positive quantity in such problems, we take the positive root:

$$v = \sqrt{64}$$ $$v = 8$$

The value of $v$ when $u = 515$ is $\boxed{\text{8}}$.