1
$\begingroup$

Question reads- Find the acute angle between the curves at their points of intersection. (The angle between two curves is the angle between their tangent lines at the point of intersection.)

$y=x^2$, $y=x^3$

This is a calculus course, unit is vectors and section is dot product.

I understand how to find the angle between two vectors. However, the derivative of $y=x^3$ (and thus the tangent line) is $y=x^2$ and I don't know how to represent that as a vector. Thanks!

  • 2
    What is (are) intersection point? Compute values of derivatives at this point. – 2017-02-04
  • 0
    Welcome to math stack exchange! Please tell us whether the given tip was helpful! – 2017-02-04
  • 0
    Could you answer to @ninjaaa ? – 2017-02-04

1 Answers 1

1

Let: $ f'(x) =(x^3)' = 3x^2$ $ g'(x) =(x^2)' = 2x$

Next solve equation:

$x^2 = x^3$

$x^2-x^3 = 0$

$x^2(1-x) = 0$

Hence intesection points are at $x=0$ and $x=1$

At $x=0$: $f'(0) = 0$ and $g'(0) $ - derivatives are equal hence angle is 0.

At $x=1$ : $f'(1)= 3$ and $g'(1)=2$ hence the angle is $\arctan(3)-\arctan(2) \approx 0.141897$ because the slope of the line equals $\tan(\alpha)$ where $\alpha$ is angle of inclination of line. (https://en.wikipedia.org/wiki/Linear_function)

  • 1
    (+1) I would just replace $arctan$ with $\arctan$ and add that $\arctan(3)-\arctan(2)=\arctan\frac{1}{7}\approx\frac{1}{7}$. – 2017-02-04
  • 0
    Yes, because $\tan(x-y) = \frac { \tan x -\tan y } {1 + \tan x \tan y }$ :) – 2017-02-04