Session 4
Lectures 12 - 17
- 14.5 The Chain Rule
- 14.6 Directional Derivatives and the Gradient Vector
- 14.7 Maximum and Minimum Values
- 14.8 Lagrange Multipliers
14.5 The Chain Rule
Case 1: Suppose \(z = f(x,y)\) is differentiable, and \(x = g(t)\), \(y = h(t)\) are differentiable functions of \(t\). Then \[
\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}
\]
Case 2: Suppose \(z = f(x,y)\) is differentiable, and \(x = g(s,t)\), \(y = h(s,t)\). Then \[
\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}
\qquad
\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}
\]
Exercises
14.5 The Chain Rule
- Let \(z = x^2 y\) where \(x = 3t\) and \(y = t^2 + 1\). Use the chain rule to find \(\frac{dz}{dt}\) when \(t = 1\)
14.6 Directional Derivatives and the Gradient Vector
The directional derivative of \(f\) at \((x_0, y_0)\) in the direction of a unit vector \(\mathbf{u} = \langle a, b \rangle\) is \[
D_{\mathbf{u}} f(x_0,y_0) = \lim_{h \to 0} \frac{f(x_0 + ha, y_0 + hb) - f(x_0,y_0)}{h}
\]
- The gradient of \(f\) is \(\nabla f(x,y) = \langle f_x(x,y), f_y(x,y) \rangle\)
- If \(f\) is differentiable, then \(D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}\)
- If the direction is given by a non-unit vector \(\mathbf{v}\), use \(\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|}\)
- \(D_{\mathbf{i}} f = f_x\) and \(D_{\mathbf{j}} f = f_y\)
14.6 Directional Derivatives and the Gradient Vector
cont.
- \(f\) increases fastest in the direction of \(\nabla f\) (at rate \(|\nabla f|\)) and decreases fastest in the direction of \(-\nabla f\)
- \(D_{\mathbf{u}} f = 0\) when \(\mathbf{u} \perp \nabla f\), i.e. along the level curve
Exercises
14.6 Directional Derivatives and the Gradient Vector
You are walking on the graph \(f(x,y) = 2y\cos(\pi x) - 2y\cos(\pi y) + 5\) and you are standing at the point \((1, 1, 4)\)
- Find a direction you should walk in order to stay at a height of \(4\). Report your answer as a unit vector
- Find a direction you should walk in order to decrease your height the fastest. Report your answer as a unit vector
14.7 Maximum and Minimum Values
- \(f\) has a local maximum at \((a,b)\) if \(f(x,y) \le f(a,b)\) for all \((x,y)\) near \((a,b)\) (similarly for local minimum)
- If \(f\) has a local extremum at \((a,b)\) and the partials exist there, then \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\)
- \((a,b)\) is a critical point if \(f_x(a,b) = f_y(a,b) = 0\), or if one of the partials does not exist
14.7 Maximum and Minimum Values
cont.
Second Derivatives Test
Suppose \((a,b)\) is a critical point and let \[
D = D(a,b) = f_{xx}(a,b)\,f_{yy}(a,b) - [f_{xy}(a,b)]^2
\]
- If \(D > 0\) and \(f_{xx}(a,b) > 0\), then \(f(a,b)\) is a local minimum
- If \(D > 0\) and \(f_{xx}(a,b) < 0\), then \(f(a,b)\) is a local maximum
- If \(D < 0\), then \((a,b)\) is a saddle point
- If \(D = 0\), the test is inconclusive
14.7 Maximum and Minimum Values
cont.
Absolute Maximum and Minimum Values
- Extreme Value Theorem: if \(f\) is continuous on a closed, bounded set \(D\), then \(f\) attains an absolute maximum and minimum on \(D\)
To find them:
- Find the values of \(f\) at the critical points of \(f\) in \(D\)
- Find the extreme values of \(f\) on the boundary of \(D\)
- The largest value from steps 1 and 2 is the absolute maximum, the smallest is the absolute minimum
Exercises
14.7 Maximum and Minimum Values
- Find and classify the critical points for the function \(h(x,y) = x^4 + y^3 - 6y - 2x^2\)
14.8 Lagrange Multipliers
To find the extreme values of \(f(x,y,z)\) subject to the constraint \(g(x,y,z) = k\) (assuming they exist and \(\nabla g \neq \mathbf{0}\) on the constraint):
- Find all \(x\), \(y\), \(z\), and \(\lambda\) such that \[
\nabla f(x,y,z) = \lambda \nabla g(x,y,z) \qquad g(x,y,z) = k
\]
- Evaluate \(f\) at all the points found in step 1. The largest is the maximum, the smallest is the minimum
- Geometrically, at an extremum the level curve/surface of \(f\) is tangent to the constraint, so \(\nabla f \parallel \nabla g\)
14.8 Lagrange Multipliers
cont.
Two Constraints
- To find extreme values of \(f(x,y,z)\) subject to \(g(x,y,z) = k\) and \(h(x,y,z) = c\), solve \[
\nabla f = \lambda \nabla g + \mu \nabla h \qquad g(x,y,z) = k \qquad h(x,y,z) = c
\]
Combining with 14.7
- To find absolute extrema on a closed region, use Lagrange multipliers to handle the boundary, and compare with the critical points in the interior
Exercises
14.8 Lagrange Multipliers
- Find the maximum and minimum values of the function \(f(x,y) = x + y\) on the curve \(x^2 + y^2 - xy = 4\)
- Consider a hypothetical planet whose surface can be approximated by the sphere \(x^2 + y^2 + z^2 = 1\). Suppose the planet’s temperature is strange and given by \(T(x,y,z) = 2x + 2y + z\). Find the maximum and minimum temperatures on the planet’s surface
Fin
This concludes session 4.