Discussion 15

Chapter 12.5

Problem 7:

In Exercises 7-12, functions \(z=f(x,y)\), \(x=g(t)\), and \(y=h(t)\) are given.

  1. Use the Multivariable Chain Rule to compute \(\frac { dz }{ dt }\).
  2. Evaluate \(\frac { dz }{ dt }\) at the indicated t-value.

In our problem we are given:

\(z=3x+4y\), \(x={ t }^{ 2 }\), \(y=2t\); \(t=1\)

Following the Multivariable Chain Rule theorem:

\(\frac { dz }{ dt } =\frac { df }{ dt } ={ f }_{ x }(x,y)\frac { dx }{ dt } +{ f }_{ y }(x,y)\frac { dy }{ dt }\)

  1. \({ f }_{ x }(x,y)=3\), \({ f }_{ y }(x,y)=4\), \(\frac { dx }{ dt } =2t\), \(\frac { dy }{ dt } =2\)

Thus,

\(\frac { dz }{ dt } =3(2t)+4(2)\)

\(\frac { dz }{ dt } =6t+8\)

  1. At \(t=1\), \(\frac { dz }{ dt } =6(1)+8=14\).