Question

Find the equation of the tangent plane to a surface\(S\)at the point\({\rm{P(2,1,3)}}.\)

Short answer

The equation of the tangent plane at\(P(2,1,3)\) is\(12x - 7y + 9z - 44 = 0\).

Step-by-step solution

Differentiate the equations.

Let\(S\)be the surface and the curves are,

\({r_1}(t) = \left\langle {2 + 3t,1 - {t^2},3 - 4t + {t^2}} \right\rangle \)

\({r_2}(u) = \left\langle {1 + {u^2},2{u^3} - 1,2u + 1} \right\rangle \)

Let\({r_1}(t) = \left\langle {2 + 3t,1 - {t^2},3 - 4t + {t^2}} \right\rangle \) …… (1)

Differentiate equation (1) with respect to\(t\)at the point\(P(2,1,3)\) and obtain the tangent vector\(r_1^\prime (t),\)

\(\begin{aligned}{l}r_1^\prime (t) = \langle 0 + 3,0 - (2t),0 - 4(1) + (2t)\rangle \\ = \langle 3, - 2t, - 4 + 2t\rangle \end{aligned}\)

\({\rm{ Let }}{r_2}(u) = \left\langle {1 + {u^2},2{u^3} - 1,2u + 1} \right\rangle \) …… (2)

Differentiate equation (2) with respect to\(u\)at the point\(P(2,1,3)\)and obtain the tangent vector\(r_2^\prime (u),\)

\(\begin{aligned}{l}r_2^\prime (u) = \langle 0\left\langle { + (2u),2\left( {3{u^2}} \right) - 0,2 + 0} \right.\\ = \left\langle {2u,6{u^2},2} \right\rangle \end{aligned}\)

Find the curves through the point.

To find the curve passes through the point\(P(2,1,3)\)or not. By assuming the value of\(t = 0\)and\(u = 1.\)

Assume\(t = 0\)and\(u = 1\)in equation (1) and (2),

\(\begin{aligned}{l}{r_1}(0) = \left\langle {2 + 3(0),1 - {{(0)}^2},3 - 4(0) + {{(0)}^2}} \right\rangle \\ = \langle 2,1,3\rangle \end{aligned}\)

Assume\(u = 1\)in equation (2),

\(\begin{aligned}{l}{r_2}(1) = \left\langle {1 + {{(1)}^2},2{{(1)}^3} - 1,2(1) + 1} \right\rangle \\ = \langle 1 + 1,2 - 1,2 + 1\rangle \\ = \langle 2,1,3\rangle \end{aligned}\)

From this, it is clear that the curve satisfies the point\(P(2,1,3)\)when\(t = 0\)and\(u = 1.\)

Substitute\(t = 0\) in the tangent vector,

\(\begin{aligned}{l}r_1^\prime (0) = \langle 3, - 2(0), - 4 + 2(0)\rangle \\ = \langle 3,0, - 4\rangle \end{aligned}\)

Substitute\(u = 1\)in the tangent vector,

\(\begin{aligned}{l}r_2^\prime (1) = \left\langle {2(1),6{{(1)}^2},2} \right\rangle \\ = \langle 2,6,2\rangle \end{aligned}\)

The tangent line is perpendicular to both\(r_1^\prime (t)\)and\(r_2^\prime (u).\)

Thus, the tangent line is parallel to the vector\(v = r_1^\prime (0) \times r_2^\prime (1).\)

Compute the values of\(v = r_1^\prime (0) \times r_2^\prime (1)\).

\(\begin{aligned}{l}r_1^\prime (0) \times r_2^\prime (1) = \left| {\begin{aligned}{*{20}{c}}i&j&k\\3&0&{ - 4}\\2&6&2\end{aligned}} \right|\\ = i(0 + 24) - j(6 + 8) + k(18 - 0)\\ = 24i - 14j + 18k\end{aligned}\)

The equation of the tangent plane is obtained as,

\(\begin{aligned}{l}24(x - 2) - 14(y - 1) + 18(z - 3) = 0\\24x - 48 - 14y + 14 + 18z - 54 = 0\\24x - 14y + 18z - 88 = 0\\12x - 7y + 9z - 44 = 0\end{aligned}\)

Therefore, the equation of the tangent plane at\(P(2,1,3)\) is\(12x - 7y + 9z - 44 = 0\).