I am given this exercise:

Find the angle between the line $y = 4 − x$ and the tangent line to the curve $y = 4 − \frac{x^2}{2}$ at the points where the line intersects the curve.

When I try it, the tangent line is $y=y_0-(x-x_0)$, so the slope is the same in both lines and the angle is 0 (they are parallel).

The answer should be $18,45º$. Where did I go wrong?

  • $\begingroup$ The tangent line where? There are lines tangent to a parabola in every direction except its opening direction. $\endgroup$ – Dan Uznanski Nov 6 '17 at 8:16
  • 1
    $\begingroup$ @DanUznanski Given that there are supposed to be those two answers, I think that the intention is to be looking at the two places where the line crosses the parabola. That's completely unclear from the question itself. $\endgroup$ – alex.jordan Nov 6 '17 at 8:22
  • $\begingroup$ That was a doubt for me too since the exercise didn't say so. $\endgroup$ – Evoked Nov 6 '17 at 18:20

You should differentiate both to get the gradient of both of them, so I'll denote the gradients as m:

$${dy_1\over dx} = -1 \\ {dy_2\over dx } = -x $$

Now you have to find where they intersect, so that happens when $y_1 = y2 $ which is at:

$$ 4 - x = 4 - {x^2 \over 2 } \\ x^2 - 2x = x(x-2) = 0$$

So at $x=0$ and $x=2$, as shown here.

So the gradients are therefore just:

$$ @ x=0 \quad m_1 = -1 \quad m_2 = 0 \\ @ x=2 \quad m_1 = -1 \quad m_2 = -2 $$

You should be able to get it from there :)


The line $y=4-x$ intersects the parabola $y=4-x^2/2$ at two places: $(0,4)$ and $(2,2)$. At the first intersection, the parabola's tangent line is $y=4$. At the second, it's $y=-2(x-2)+2$.

So you need to find the angle between slope $-1$ and slope $0$. And then the angle between slope $-1$ and slope $-2$.

The first angle is $\arctan(0)-\arctan(-1)$, which is $45^{\circ}$.

For the second you need, $\arctan(-1)-\arctan(-2)$ which rounds to the stated answer of $18^{\circ}$, but is not exactly $18^{\circ}$.


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