Thread: How to find parametric equations for the three level curves

Find parametric equations for the three level curves of the function
W(x,y) = sin(x) e^y
which pass through the points P = (0,1), Q = (pi/2, 0) and R = (pi/6, 3)
Also compute the vectors of the gradient vector field (gradient of W) at the points P, Q an R
Is this correct to find gradient as follows?
Find the partial derivatives dW/dx and dW/dy
Then find (dW/dx) i + (dW/dy) j at the three points.
But I do not know how to find the parametric equations.

Do you know what the functions are? Do you know what "level curves" are? If we are talking about a level curve of [tex]W(x, y)= sin(x)e^{y}[/itex] that contains (0, 1), what equation must be true? All of these questions are pretty close to trivial if you know the definition of "level curve".

Originally Posted by visharad

Find parametric equations for the three level curves of the function W(x,y) = sin(x) e^y
which pass through the points P = (0,1)

$\displaystyle W(0,1)=0,$ so the level set is $\displaystyle (\sin x)e^y=0$ or equivalently $\displaystyle \sin x=0$ and $\displaystyle y\in\mathbb{R}.$ As a consequence we get the family of curves $\displaystyle C_k:\begin{Bmatrix} x=k\pi\\y=t\end{matrix} \quad(t\in\mathbb{R})$ where $\displaystyle k\in\mathbb{Z}.$

Q = (pi/2, 0)

Now, $\displaystyle W(\pi/2,0)=1$ so, the level set is $\displaystyle (\sin x) e^y=1.$ Express $\displaystyle e^y=\csc x$ or $\displaystyle y=\ln (\csc x)$ and denote $\displaystyle x=t.$


Edited: Sorry, I didn't see HallsofIvy's post.

Hi visharad.

You're correct in your description for finding the gradient then evaluating it at the given points, nice work! I'll give you some thoughts on finding the parametric equation for the level curve through P=(0,1), then let you take another shot at the other two.

To start, a level curve is the set of point in the domain (in this case $\displaystyle \mathbb{R}^{2}$) where the function is constant. At P=(0,1) we have $\displaystyle W(0,1)=\sin(0)e^{1}=0.$ This means we want to look for all points in the domain where $\displaystyle W$ is zero. This set is given by

$\displaystyle S=\{(x,y)\in \mathbb{R}^{2}: W(x,y)=0\}$

Setting up the equation $\displaystyle W(x,y)=0$ gives $\displaystyle \sin xe^{y}=0.$ Since $\displaystyle e^{y}\neq 0$ for all $\displaystyle y$, we can divide by $\displaystyle e^{y}$ to get $\displaystyle \sin x=0.$ Now, $\displaystyle \sin x=0$ when $\displaystyle x=n\pi$ where $\displaystyle n\in \mathbb{Z}.$ Hence, the set S above giving the level curves for $\displaystyle W(x,y)=0$ can be written

$\displaystyle S=\{(x,y)\in \mathbb{R}^{2}: x=n\pi, n\in\mathbb{Z}\}.$

Geometrically, S is a collection of vertical lines in the plane; i.e. there are vetrical lines at $\displaystyle x=0, x=\pm \pi, x=\pm 2\pi, $ etc. The level curve passing through (0,1) is the vertical line x=0. This can be parameterized by the function $\displaystyle \alpha(t)=(0, t),$ where $\displaystyle t\in\mathbb{R}.$

Does this help? Let me know if you have any questions.

Good luck!

z=w(x,y) as a surface:
z=0 plane also displayed:



Contour lines of z=w(x,y)



z=w(x,y)=c only produces parallel lines

At Q=(pi/2,0): W(pi/2,0) = sin(pi/2) e^0 = 1
sin(x) e^y = 1
=> e^y = cscx
=> y = ln(cscx)

For the above to be valid, csc(x) > 0
=> 2npi < x < (2n+1)pi
$\displaystyle S = \{(x,y)\in\mathbb{R}^{2}: 2n\pi < x < (2n+1)\pi, n \in\mathbb{Z}, y = ln(cscx)\}$

Similarly, at R we get
$\displaystyle S = \{(x,y)\in\mathbb{R}^{2}: 2n\pi < x < (2n+1)\pi, n \in\mathbb{Z}, y = ln((1/2) e^3 cscx)\}$

Is this correct? And what about gradient part?