Question

Determine the derivative \({W_s}(1,0)\) and\({W_t}(1,0)\). The functions are \(z = f(x,y),x = g(t)\) and \(y = h(t){\rm{. }}\)

Short answer

The value of derivative are \({W_s}(1,0) = 52\) and \({W_t}(1,0) = 34.\)

Step-by-step solution

The chain rule.

With the chain rule, the partial derivative of function\({\rm{f(x, y)}}\)with respect to\({\rm{t}}\)becomes\(\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.\)

Use the chain rule for calculation.

The given functions are \(z = f(x,y),x = g(t)\) and \(y = h(t){\rm{. }}\)

Differentiate \(W\) partially with respect to \(s\)as follows.

\(\begin{aligned}{l}\frac{{\partial W}}{{\partial s}} = \frac{\partial }{{\partial s}}F(u(s,t),v(s,t))\\ = \frac{{\partial F}}{{\partial u}} \cdot \frac{{\partial u}}{{\partial s}} + \frac{{\partial F}}{{\partial v}} \cdot \frac{{\partial v}}{{\partial s}}\\ = {F_u}(u(s,t),v(s,t)) \cdot {u_s}(s,t) + {F_v}(u(s,t),v(s,t)) \cdot {v_s}(s,t){\rm{ }}\end{aligned}\)

At \(s = 1\) and\(t = 0\), the partial derivative becomes.

\(\begin{aligned}{l}{\left. {\frac{{\partial W}}{{\partial s}}} \right|_{s = 1,t = 0}} = {F_u}(u(1,0),v(1,0)) \cdot {u_s}(1,0) + {F_v}(u(1,0),v(1,0)) \cdot {v_s}(1,0)\\ = {F_u}(2,3) \cdot {u_s}(1,0) + {F_v}(2,3) \cdot {v_s}(1,0)\\ = ( - 1) \cdot ( - 2) + (10) \cdot (5)\\ = 2 + 50\end{aligned}\)

That is, \({\rm{Ws(1,0) = 52}}{\rm{.}}\)

Differentiate \(W\) partially with respect to \(t\)as follows.

\(\begin{aligned}{l}\frac{{\partial W}}{{\partial t}} = \frac{\partial }{{\partial t}}F(u(s,t),v(s,t))\\ = \frac{{\partial F}}{{\partial u}} \cdot \frac{{\partial u}}{{\partial t}} + \frac{{\partial F}}{{\partial v}} \cdot \frac{{\partial v}}{{\partial t}}\\ = {F_u}(u(s,t),v(s,t)) \cdot {u_t}(s,t) + {F_v}(u(s,t),v(s,t)) \cdot {v_t}(s,t){\rm{ }}\end{aligned}\)

At \(s = 1\) and\(t = 0\), the partial derivative becomes.

\(\begin{aligned}{l}{\left. {\frac{{\partial W}}{{\partial t}}} \right|_{s = 1,t = 0}} = {F_u}(u(1,0),v(1,0)) \cdot {u_t}(1,0) + {F_v}(u(1,0),v(1,0)) \cdot {v_t}(1,0)\\ = {F_u}(2,3) \cdot {u_t}(1,0) + {F_v}(2,3) \cdot {v_t}(1,0)\\ = ( - 1) \cdot (6) + (10) \cdot (4)\\ = - 6 + 40\end{aligned}\)

That is, \({\rm{Wt(1,0) = 34}}{\rm{.}}\)

Therefore, the value of derivative are \({W_s}(1,0) = 52\) and \({W_t}(1,0) = 34.\)