Question

(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)

Answer 1

\[x^{3} = \sqrt[3]{x}\]

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

Answer 2

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

Answer 3

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?

Answer 4

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

Answer 5

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
\[\cos(\theta) = \frac{6}{10}\]
\[\theta =53.13\]

Answer 6

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

Answer 7

thankss!