calculusvectors
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!
Eliminating the parameter $t$ from the first equation yields
$$x=(y-1)^2 $$
giving
$$\frac{dx}{dy}=2(y-1)\text{ or }\frac{dy}{dx}=\frac{1}{2(y-1)}$$
which gives $\dfrac{dx}{dy}$ a value $0$ at $(0,1)$. So the first curve has a vertical tangent at $(0,1)$ and a slope of $\dfrac{dy}{dx}=-\frac{1}{2}$ at $(0,1)$.
For the second equation we may let
$$ F(x,y)=5x^2+5xy+3y^2-8x-6y+3 $$
and find the implicit derivative using the partial derivative identity
$$ \frac{dy}{dx}=-\frac{F_x}{F_y} $$
to obtain
$$ \frac{dy}{dx}=-\frac{10x+5y-8}{5x+6y-6} $$
alternately
$$ \frac{dx}{dy}=-\frac{5x+6y-6}{10x+5y-8} $$
This shows that at the point $(1,0)$ we have
$$ \frac{dy}{dx}=2 $$
and at the point $(0,1)$ we have a vertical tangent since $\dfrac{dx}{dy}=0$ at that point.
Thus at the point $(0,1)$ both curves have a vertical tangent. So the angle between their tangents is $0$ at $(0,1)$.
At the point $(1,0)$ the slopes are the negative reciprocal of each other so the angle between their tangents is $90^\circ$ or $\pi/2$ radians at that point.
Best Answer
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)