This is a solved example from Calculus Made Easy (you can download it from the link), I have the solution but can't understand it.

  • PROBLEM 1: How did author find the angle ?
  • PROBLEM 2: Author insists that it is tangent, but graph shows it's secant.
  • PROBLEM 3: Page 86, author writes: Now, when two curves meet, the intersection being a point common to both curves, its coordinates must satisfy the equation of each one of the two curves; He is talking about intersection of 2 curves here but equations he equates are of tangent and one curve. It is confusing.

STATEMENT: Find the slope of the tangent to the curve $y = \frac{1}{2x} + 3$ at $x = -1$. Find the angle this tangent makes with 2nd curve $y = 2x^2 + 2$.

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SOLUTION: at $x = -1,$ Slope of tangent to curve = Slope of the curve.

  1. $\frac{dy}{dx}$ = $\frac{-1}{2x^2}$, putting x = -1, $\frac{dy}{dx}$ = $\frac{-1}{2}$.

  2. Tangent is a straight line, $y = ax + b$, which means $\frac{dy}{dx}$ = $a$ = $-1/2$

  3. For first curve, $x = -1$ means $y = \frac{5}{2}$.

  4. We got values for $x$, $y$ and $a$. Since tangent touches and curve touch at one point, co-ordinates of that point will be same for both tangent and curve must be same. To find $b$ we put put all these values in equation for tangent $y = ax + b$, which gives us $b = 2$. hence equation of tangent is $y = \frac{-x}{2} +2$.

  5. From (4). equation for first curve = equation for tangent . $2x^2 + 2 = \frac{-x}{2} + 2$. Solving this we get $x = 0$ or $x = \frac14$

Now we will go to 2nd curve

  1. $y = 2x^2 +2$, $\frac{dy}{dx} = 4x$ and for $x = \frac{-1}{4}$, $\frac{dy}{dx} = -1$. Since tan(45) = 1. Hence the slope for this 2nd curve is downwards.

  2. Slope for the tangent is -1/2, hence that is downwards too with $tan(angle) = \frac{1}{2}$.

I do not understand from here onwards

it slopes downwards to the right and makes with the horizontal an angle φ such that tan φ = 12 ; that is, an angle of 26 degrees 34 minutes. It follows that at the first point the curve cuts the straight line at an angle of 26 degrees 34 minutes, while at the second it cuts it at an angle of 45 degrees − 26 degrees 34 minutes = 18 degrees 26 minutes. Hence the answer is 18 degrees 26 minutes

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    $\begingroup$ The question: "Find the angle this tangent makes with 2nd curve $y = 2x^2 + 2$" is not much clear to me. $\endgroup$ – user Aug 8 '18 at 13:11
  • $\begingroup$ The answer, 18 deg 26 min is the acute angle that your chord makes with a tangent line to the 2nd curve at the point x=(-1/4). $\endgroup$ – Narlin Aug 8 '18 at 13:25
  • $\begingroup$ @Narlin , my chord and tangent to 2nd curve. I don't understand, there is only 1 line, that is one tangent. From where chord came ? $\endgroup$ – Arnuld Aug 8 '18 at 13:57
  • $\begingroup$ @gimusi -- tangent line touches both of the curves, 1st curve at 2 points and 2nd curve at 1 point, hence total 3 points. Just like you I can't make out either that which point's angle author is asking about. $\endgroup$ – Arnuld Aug 8 '18 at 14:02
  • $\begingroup$ @Arnuld Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Sep 6 '18 at 23:13

"Find the angle this tangent makes with 2nd curve" Since the tangent line intersects the curve $y=2x^2+2$, it must make some angle with that curve at that spot (the spot is y(x=-1/4). So you find a new tangent to the second curve, which is $y=-x+15/8$. Then you get the acute angle between these two lines.enter image description here

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  • $\begingroup$ Thanks Narlin. I understood what you explained here. It is still half the story. I could never understand on what basis author did (45 - 26.34) to get the answer. $\endgroup$ – Arnuld Aug 8 '18 at 14:29
  • $\begingroup$ Yeah! It is unusual. I wouldn't bother. We understand the question. We understand a way to get the answer. Done. $\endgroup$ – Narlin Aug 8 '18 at 14:33

There is something wrong here

  1. From (4). equation for first curve = equation for tangent . $2x^2 + 2 = \frac{-x}{2} + 2$. Solving this we get $x = 0$ or $x = 4$

we should have

$$2x^2 + 2 = \frac{-x}{2} + 2 \iff4x^2+x=0 \iff x=0 \quad x=-\frac14$$

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  • $\begingroup$ You have written exactly the same equation. Only thing different is: your MathJax is good $\endgroup$ – Arnuld Aug 8 '18 at 14:12
  • $\begingroup$ @Arnuld it's not the same, we obtain $x=0$ and $x=-1/4$ and not $x=4$. $\endgroup$ – user Aug 8 '18 at 14:26
  • $\begingroup$ Left Hand Side (LHS) of your equation says "(-x/2) +2" and it is same as in my post "(-x/2) +2". That looks same to me $\endgroup$ – Arnuld Aug 8 '18 at 14:31
  • $\begingroup$ @Arnuld Yes of course I've copied that. But your result seems to be wrong. $\endgroup$ – user Aug 8 '18 at 14:33
  • $\begingroup$ Oh my bad, Mathjax messed up my mind :-\ . It was a typo I see. Thank you for your patience. I will correct the OP $\endgroup$ – Arnuld Aug 8 '18 at 15:57

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