Finding the x coordinates where a curve is tangent to a line passing through a point

[xdxdxdxd]

Find the x-coordinates of all the points on the following curve:

. . . . .\(\displaystyle y\, =\, 12x\, \cos(17x)\, -\, 102\, \sqrt{\strut 2\,}\, x^2\, -\, 46\)

...on the following interval:

. . . . .\(\displaystyle \dfrac{\pi}{17}\, <\, x\, <\, \dfrac{2\pi}{17}\)

...where the tangent line passes through the following point P (which is not on the curve):

. . . . .\(\displaystyle P\, =\, (0,\, -46)\)

---

After taking the derivative (m), I equated it to (y - y₁)/(x - x₁) using (-46) as my y₁, (0) as x₁, and the given equation as y. This ultimately left me with 204*SQRT(2)*sin(17x)+102*SQRT(2)*x^2 = 0. I have no idea how to solve for x from here.

 

[tkhunny]

Find the x-coordinates of all the points on the following curve:

. . . . .\(\displaystyle y\, =\, 12x\, \cos(17x)\, -\, 102\, \sqrt{\strut 2\,}\, x^2\, -\, 46\)

...on the following interval:

. . . . .\(\displaystyle \dfrac{\pi}{17}\, <\, x\, <\, \dfrac{2\pi}{17}\)

...where the tangent line passes through the following point P (which is not on the curve):

. . . . .\(\displaystyle P\, =\, (0,\, -46)\)

---

After taking the derivative (m), I equated it to (y - y₁)/(x - x₁) using (-46) as my y₁, (0) as x₁, and the given equation as y. This ultimately left me with 204*SQRT(2)*sin(17x)+102*SQRT(2)*x^2 = 0. I have no idea how to solve for x from here.

You can't.

How did (0,-46) get off the curve?

 

[xdxdxdxd]

You can't.

How did (0,-46) get off the curve?

What do you mean how? The question just defines the point as a point exclusive to the straight line equations and not on the curve.

 

[gregk]

What do you mean how? The question just defines the point as a point exclusive to the straight line equations and not on the curve.

But \(\displaystyle (0,-46)\) is clearly on the given curve. So something isn't right.

Edit: never mind, I didn't pay attention to the explicit domain there.

 

[gregk]

So we start with

\(\displaystyle y = 12x\cos(17x) - 102\sqrt2x^2 - 46, \quad \dfrac\pi{17} < x < \dfrac{2\pi}{17}.\)

Taking the derivative and simplifying, we get

\(\displaystyle y' = 12\cos(17x) - 204x\sin(17x) - 204\sqrt2x.\)

What we want is a tangent line that passes through \(\displaystyle (0,-46)\). Such a line would have the form

\(\displaystyle y = mx - 46\)

If this is to be a tangent line, then the line must pass through a point \(\displaystyle (x,y)\) on the given curve, and the line must have a slope of \(\displaystyle y'(x)\). Substituting for \(\displaystyle y\) and \(\displaystyle y'\) gives

\(\displaystyle
\left(12x\cos(17x) - 102\sqrt2x^2 - 46\right) = \left(12\cos(17x) - 204x\sin(17x) - 204\sqrt2x\right)x - 46.
\)

Now a bunch of stuff will cancel out, and after rearranging and simplifying you'll have a simple sine equation. Can you take it from here?