The equation of the tangent to the curve $\mathrm{x}^{\frac{5}{2}}+\mathrm{y}^{\frac{5}{2}}=33$ at the point $(1,4)$ is :

Verify that $(1, 4)$ lies on the curve:

$1^{\frac{5}{2}} + 4^{\frac{5}{2}} = 1 + (\sqrt{4})^5 = 1 + 32 = 33 \checkmark$

Differentiate both sides with respect to $x$ using implicit differentiation:

$\dfrac{5}{2}\,x^{\frac{3}{2}} + \dfrac{5}{2}\,y^{\frac{3}{2}} \cdot \dfrac{dy}{dx} = 0$

Since $y$ is a function of $x$, the chain rule gives us the $\dfrac{dy}{dx}$ term on the second part.

Solve for $\dfrac{dy}{dx}$. The $\dfrac{5}{2}$ cancels from both sides:

$\dfrac{dy}{dx} = -\dfrac{x^{\frac{3}{2}}}{y^{\frac{3}{2}}}$

Substitute the point $(1, 4)$:

$\dfrac{dy}{dx}\bigg|_{(1,4)} = -\left(\dfrac{1}{4}\right)^{\frac{3}{2}}$

$= -\left(\sqrt{\dfrac{1}{4}}\right)^3 = -\left(\dfrac{1}{2}\right)^3 = -\dfrac{1}{8}$

Using point-slope form $y - y_1 = m(x - x_1)$:

$y - 4 = -\dfrac{1}{8}(x - 1)$

Multiply both sides by $8$:

$8y - 32 = -x + 1$

$x + 8y = 33$

$\boxed{x + 8y = 33}$