Question

Find the equation of tangents to the curve
y=cos(x+y), -2π≤x≤2π,
that are parallel to the line x+2y=0?

Answer 1

Have you considered the 1st derivative?

Answer 2

oh ..yes..
but the tangents that come are really not tangent when i draw them up in geogebra

Answer 3

You shouldn't have to rely so much on graphing software to find an answer.

Differentiating gives
\[\begin{align*}
y&=\cos(x+y)\\
\frac{dy}{dx}&=-\sin(x+y)\left(1+\frac{dy}{dx}\right)\\
\frac{dy}{dx}&=-\sin(x+y)-\sin(x+y)\frac{dy}{dx}\\
\frac{dy}{dx}\left(1+\sin(x+y)\right)&=-\sin(x+y)\\
\frac{dy}{dx}&=-\frac{\sin(x+y)}{1+\sin(x+y)}
\end{align*}\]
The slope of the given line \(x+2y=0\) is \(-\dfrac{1}{2}\), so you want to find all the tangent lines with the same slope. Set equal to the derivative:
\[-\frac{1}{2}=-\frac{\sin(x+y)}{1+\sin(x+y)}\]
This tells you that \(\sin(x+y)=1\), which occurs for \(x+y=\dfrac{\pi}{2}+2n\pi\) (for \(n=0,-1\), or \(\dfrac{\pi}{2}\) and \(-\dfrac{3\pi}{2}\).

To actually find these points, you could try something like this:
\[y=\cos(x+y)~~\Rightarrow~~y=\cos\left(\frac{\pi}{2}+2n\pi\right)=0\]
So it seems that any point on the curve with a \(y\)-coordinate of 0 will give you a tangent line with slope \(-\dfrac{1}{2}\).
\[-\frac{1}{2}=-\frac{\sin(x+0)}{1+\sin(x+0)}~~\Rightarrow~~\sin x=1~~\Rightarrow~~x=\frac{\pi}{2},-\frac{3\pi}{2}\]
So you have two tangent lines, tangent to \(\left(\dfrac{\pi}{2},0\right)\) and \(\left(-\dfrac{3\pi}{2},0\right)\). You know the slopes are \(-\dfrac{1}{2}\). You thus have everything you need to figure out the tangent line equations.

Answer 4

The assumption in the implicit derivative, dy/dx, is that locally, y = f(x). You do need to be somewhat concerned about where this is not the case.

Anywhere the slope is vertical violates this assumption. Since we are looking for a slope of 1/2, this shouldn't be a problem.

Anywhere the curve crosses itself violates this assumption. This will take more care.

Answer 5

True... wouldn't it be enough to determine where \(\dfrac{dx}{dy}=0\) and restrict the domain?

Answer 6

At any rate, the plot on Wolfram seems to agree that \(\left(\dfrac{\pi}{2},0\right)\) and \(\left(-\dfrac{3\pi}{2},0\right)\) are okay, but the site's implicit plots can be sketchy at times.

Answer 7

Yes, the vertical is solved by assuming x = f(y).

The intersections take greater care.

Answer 8

@SithsAndGiggles - i already had the solution...but when i draw them in geogebra, why dont they just touch the curve instead of cutting them..

Answer 9

Because a tangent line doesn't have to not intersect the curve again.

Consider the function \(y=x^3\). You can have a tangent that intersect the curve at least one other time: