Chapter 14: Problem 58
Normal curves \(A\) smooth curve is normal to a surface \(f(x, y, z)=c\) at a point of intersection if the curve's velocity vector is a nonzero scalar multiple of \(\nabla f\) at the point. Show that the curve $$ \mathbf{r}(t)=\sqrt{t} \mathbf{i}+\sqrt{t} \mathbf{j}-\frac{1}{4}(t+3) \mathbf{k} $$ is normal to the surface \(x^{2}+y^{2}-z=3\) when \(t=1\)
Short Answer
Step by step solution
Calculate the gradient of the surface
Determine the position vector at t=1
Compute the velocity vector of the curve
Evaluate the gradient at the point of intersection
Check if the velocity vector is a scalar multiple of the gradient
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Vector
- The components of \(abla f\) reveal the function's rate of increase in the direction of each coordinate axis.
- For our example surface \(f(x, y, z) = x^2 + y^2 - z\), the gradient becomes \((2x, 2y, -1)\).
Velocity Vector
For instance, \(\mathbf{r}(t) = \sqrt{t} \mathbf{i} + \sqrt{t} \mathbf{j} - \frac{1}{4}(t+3) \mathbf{k}\) has the velocity vector:\[\mathbf{v}(t) = \frac{d}{dt}\left(\sqrt{t} \mathbf{i} + \sqrt{t} \mathbf{j} - \frac{1}{4}(t+3) \mathbf{k}\right) = \frac{1}{2\sqrt{t}} \mathbf{i} + \frac{1}{2\sqrt{t}} \mathbf{j} - \frac{1}{4} \mathbf{k}\]
- At \(t = 1\), the velocity vector becomes \(\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} - \frac{1}{4} \mathbf{k}\).
- This vector indicates how the point moves around on the curve at \(t = 1\).
Surface Intersection
- The intersection in the context of a function \(x^2 + y^2 - z = 3\) and curve \(\mathbf{r}(t)\) occurs at specific points calculated by substituting \(t\) values.
- The point found by setting \(t = 1\) and substituting into \(\mathbf{r}(t)\) gives \((1, 1, -1)\).
If the velocity vector at this point turns out to be a scalar multiple of the gradient vector, then they are pointing in the same direction, proving that the curve is normal to the surface. Checking component-wise consistency, as shown with scalar \(k\), strengthens this conclusion.