at what points is tangent parallel to curve

[fran1942]

Hello, I am trying to solve the following question:
At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.

1. rewrite 3x-y=5 as y-3x-5
2. equate 2(x-cosx) = y-3x-5
3. differentiate: 2+2sinx = 3
4. solve for x: sin^-1(,5) = 0.524
5. My answer would be "x=pi/6 +2pi*n" and y=????????

So, as this is a trig equation I would have an infinite number of answers as per my answer above in 5.
My problem is I am not sure how I should express the y values in order to answer the original question "At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5".

Any help would be much appreciated.

 

[galactus]

Sorry, I tried. Apparently, the LaTex is down. <img src="images/smilies/icon_sad.gif" alt="" title="" smilieid="11" class="inlineimg" border="0"><br>
<br>
\(\displaystyle f'(x)=2(sin(x)+2)\)<br>
<br>
Setting this equal to 3 and solving for x we get:<br>
<br>
\(\displaystyle x=2\pi C+\frac{\pi}{6}, \;\ x=2\pi C+\frac{5\pi}{6}\)<br>
<br>
So, for the general form, try using \(\displaystyle y=mx+b\). We know the slope is m = 3.<br>
<br>
Any x value can be obtained by using integer values of C.<br>
<br>
Subbing the above general x solutions back into f(x) we get:<br>
<br>
For \(\displaystyle x=2C\pi +\frac{\pi}{6}\):<br>
<br>
\(\displaystyle y= 3x-\frac{12cos(2C\pi +\frac{\pi}{6})+\pi (12C+1)}{6}\)<br>
<br>
For \(\displaystyle x=2C\pi +\frac{5\pi}{6}\):<br>
<br>
\(\displaystyle y = 3x+\frac{12sin(2C\pi +\frac{\pi}{3})-\pi(12C+5)}{6}\)<br>
<br>
i.e. for the first one let C=0, 1, 2,...... and we get lines of the form:<br>
<br>
\(\displaystyle y = 3x-\sqrt{3}-\frac{\pi}{6}\)<br>
<br>
\(\displaystyle y = 3x-\sqrt{3}=\frac{13\pi}{6}\)<br>
<br>
.<br>
.<br>
.<br>
.<br>
\(\displaystyle y =3x- \sqrt{3}-(2C\pi +\frac{\pi}{6}\)<br>
<br>
For the other set with \(\displaystyle \frac{5\pi}{6}\) we get lines of the form:<br>
<br>
\(\displaystyle y=3x+\sqrt{3}-\frac{5\pi}{6}\)<br>
<br>
\(\displaystyle y = 3x+\sqrt{3}-\frac{17\pi}{6}\)<br>
<br>
.<br>
.<br>
.<br>
.<br>
<br>
\(\displaystyle y = 3x+\sqrt{3}-(2C\pi +\frac{5\pi}{6})\)

 

[wjm11]

At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.

Start by rearranging the linear equation:

3x-y=5
y = 3x – 5

Therefore, the slope is 3. Your problem can now be restated as “At what point(s) on the curve y=2(x-cosx) does the tangent equal 3.”

Next, find the derivative:

y = 2(x-cosx)
dy/dx = 2 + 2sinx

Set the derivative equal to the desired slope. Solve.

3 = 2 + 2sinx
sinx = ½
arcsin(sinx) = arcsin(1/2)
x = pi/6 (for the 1^st quadrant)

There is also a solution in the 2^nd quadrant. Find it.

There are an infinite number of solutions for x based on these two solutions.

Those are your x values. Sub them back into your original equation to get the corresponding y values.

 

[HallsofIvy]

Hello, I am trying to solve the following question:
At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.

1. rewrite 3x-y=5 as y-3x-5
2. equate 2(x-cosx) = y-3x-5

No. The condition is that the tangent be parallel to 2x- y= 5, not that it be the same line. There is no requirement that the line touch the curve.

3. differentiate: 2+2sinx = 3
4. solve for x: sin^-1(,5) = 0.524
5. My answer would be "x=pi/6 +2pi*n" and y=????????

So, as this is a trig equation I would have an infinite number of answers as per my answer above in 5.
My problem is I am not sure how I should express the y values in order to answer the original question "At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5".

Any help would be much appreciated.