Let $ \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} $. Then \begin{align*} \|\mathbf{A} \mathbf{v}\| &= \left\| \begin{pmatrix} 4 & 7 \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right\| \\ &= \left\| \begin{pmatrix} 4x + 7y \\ cx + dy \end{pmatrix} \right\| \\ &= \sqrt{(4x + 7y)^2 + (cx + dy)^2} \\ &= \sqrt{16x^2 + 56y + 49y^2 + c^2 x^2 + 2cd xy + d^2 y^2} \\ &= \sqrt{(c^2 + 16) x^2 + (2cd + 56) xy + (d^2 + 49) y^2}.\end{align*}We are told that we can find this value given the value of $ \|\mathbf{v}\| = \sqrt{x^2 + y^2} $. This holds if and only if $ c^2 + 16 = d^2 + 49 $ and $ 2cd + 56 = 0 $. This gives us $ c^2 - d^2 = 33 $ and $ cd = -28 $. Squaring $ c^2 - d^2 = 33, $ we get \[c^4 - 2c^2 d^2 + d^4 = 1089.\]Then \[c^4 + 2c^2 d^2 + d^4 = 1089 + 4c^2 d^2 = 1089 + 4 \cdot (-28)^2 = 4225.\]Thus, $ (c^2 + d^2)^2 = 4225 $. Since $ c^2 + d^2 $ must be nonnegative, $ c^2 + d^2 = \sqrt{4225} = 65 $. Then \[c^2 - 2cd + d^2 = 65 - 2(-28) = 121,\]so $ |c - d| = \boxed{11} $. With some more work, we can show that $ (c,d) $ is either $ (7,-4) $ or $ (-7,4) $.