## anonymous 4 years ago (a) find all points of intersection of the graphs of the two equations, (b) find the unit tangent vectors to each curve at their points of intersection, and (c) find the angles between the curves at their points of intersection. y = x^3 and y = x^(1/3)

1. dumbcow

$x^{3} = \sqrt[3]{x}$ $x(x^{8}-1)=0$ $\rightarrow x = -1,0,1$

2. anonymous

ooh thank you. do you have any idea how to do part B or C? :T

3. dumbcow

well im not sure exactly what they are looking for to find the slope tangent to each curve, take the derivative and do they want the angle between the vectors?

4. anonymous

they're looking for the angle between the vectors i think

5. dumbcow

after taking derivative of each function $dy/dx = 3x^{2}$ $dy/dx = \frac{1}{3\sqrt[3]{x}^{2}}$ at points x=+-1, function A has slope 3 function B has slope 1/3 in vector form: function A : <1,3> function B : <3,1> Both have same magnitude sqrt(3^2 +1) = sqrt10 unit vectors: <1/sqrt10, 3/sqrt10> <3/sqrt10, 1/sqrt10> to find angle use : $\cos(\theta) = \frac{u*v}{|u|*|v|}$ u*v = 1*3 + 3*1 = 6 |u|=|v| =sqrt10 $\cos(\theta) = \frac{6}{10}$ $\theta =53.13$

6. dumbcow

oh at x=0, slopes are 0 and undefined thus angle is 90

7. anonymous

thankss!