Math 280 homework notes

• Section 10.1 #26
  1. You seem to have deduced this result geometrically (which is fine). You should also be able to deduce or prove this algebraically.
  2. You have found a specific point that satisfies the given condition. There are many others. You should fit the set of all points that satisfy the condition.
• Section 10.1 #34
  1. Be careful with the directions of these inequalities. On and outside the sphere of radius 1 is given by \(x^2+y^2+z^2\geq 1\). On and inside the sphere of radius 2 is given by \(x^2+y^2+z^2\leq 4\).
  2. The two separate inequalities can also be written in one string as \(1\leq x^2+y^2+z^2\leq 4\).
• Section 10.6 #42
  1. The \(y=0\) cross-section is empty. For values \(y_0 <-1\) and \(y_0>1\), each \(y=y_0\) cross-section is a circle.
  2. A more complete description might be hyperboloid of two sheets with central axis on the y-axis and circular cross-sections perpendicular to that central axis.
• Section 12.5 #18

  The gradient vector is \(\vec\nabla f(1,0)=2\hat\jmath\). So, \(f\) increases most rapidly in the direction \(\hat\jmath\). The derivative in this direction is 2. We can express this as \[ \Bigl(\frac{df}{ds}\Bigr)_{\hat\jmath}=2 \qquad\textrm{or}\qquad D_{\hat\jmath}f(1,0)=2. \] Also, \(f\) decreases most rapidly in the direction \(-\hat\jmath\) with the derivative in this direction being \(-2\). We can express this as \[ \Bigl(\frac{df}{ds}\Bigr)_{-\hat\jmath}=-2 \qquad\textrm{or}\qquad D_{-\hat\jmath}f(1,0)=-2. \]

• Section 12.5 #32

  At the point \(P\), the gradient vector \(\vec\nabla f(P)\) has the same direction as the given vector \(\vec{v}=\hat\imath+\hat\jmath-\hat k\) and has the given magnitude \(2\sqrt{3}\). So, we can write \[ \vec\nabla f(P) =(\textrm{magnitude})(\textrm{direction}) =(2\sqrt{3})\Bigl(\frac{\hat\imath+\hat\jmath-\hat k}{\sqrt{3}}\Bigl) =2\hat\imath+2\hat\jmath-2\hat k \] To find the derivative in the direction of \(\hat\imath+\hat\jmath\), we compute \[ \Bigl(\frac{df}{ds}\Bigr)_{\hat u}=\hat{u}\cdot\vec\nabla f(P) =\frac{\hat\imath+\hat\jmath}{\sqrt{2}}\cdot \bigl(2\hat\imath+2\hat\jmath-2\hat k\bigr) =\dots=2\sqrt{2}. \] Note that this directional derivative is less than \(2\sqrt{3}\) as it must be since \(2\sqrt{3}\) is the greatest rate of change.