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!
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$\begingroup$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. ()
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