help me out with Calculus!
The angle between two differentiable curves at a point of intersection is the angle between the curves' tangent lines at this point. Find the angle between the curves in (a) and (b) at their points of intersection.
(a) y= 1-x^2, y=x^2 (two points of intersection)
(b) y=x^2, x=y^3 (two points of intersection)

Question

help me out with Calculus! 
The angle between two differentiable curves at a point of intersection is the angle between the curves' tangent lines at this point. Find the angle between the curves in (a) and (b) at their points of intersection.
(a) y= 1-x^2,  y=x^2  (two points of intersection)
(b) y=x^2,   x=y^3  (two points of intersection)

experimentx · Answer

1) solve and find the points of intersection first. http://www.wolframalpha.com/input/?i=y%3D+1-x^2%2C+y%3Dx^2+ 2) find the tangents, (dy/dx) at that point gives the slope of tangent 3) find the angle ..

anonymous · Answer

can you please explain more?

experimentx · Answer

which step??

anonymous · Answer

2nd and 3rd

experimentx · Answer

find dy/dx of both,
y= 1-x^2, y=x^2 

dy/dx = -2x, dy/dx = 2x

you know the values of x and y from step 1) right??

put those values, on above, you will have two slopes of tangents, m1 and m3


You know this formula right,
$$ \tan(\theta) = \frac{m1 - m2}{1 + m1*m2}$$

put those values of m1 and m2 and find the angle. And since it's difference, always take the positive value ... or negative won't affect it either.

anonymous · Answer

there are two points of intersection, so im not sure how i can only get two slopes of tangents

anonymous · Answer

can someone explain this

experimentx · Answer

well, find the slopes of tangent at one point of intersection... there you will have the angle between two curves.

Then move to the other point of intersection, there you will have another angle of intersection.

Good luck ...!!! it's going to lot of work!!!

anonymous · Answer

so im supposed to have 4 angles for part a?....

anonymous · Answer

can you teach me part a step by step and ill do part b.

amistre64 · Answer

the points of intersection are when y = y

amistre64 · Answer

y= 1-x^2,  y=x^2

y = y

1-x^2 = x^2  and solve for x

anonymous · Answer

yeah i got the answer for that part, and then?:)

amistre64 · Answer

what are the points of intersection?  we will need them

amistre64 · Answer

in calculus we learn that the derivative of an equation defines the slope of the tangent line at any given point

anonymous · Answer

$$(-1/\sqrt{2} , 1/2) , (1/\sqrt{2} ,1/2)$$

amistre64 · Answer

goo, the x parts are all we will need inthis one since thats all we have to work with in the equations

amistre64 · Answer

good, not goo lol

anonymous · Answer

haha

amistre64 · Answer

what are our derivatives?

anonymous · Answer

dy/dx = -2x and dy/dx=2x

amistre64 · Answer

good, plug in x = 1/sqrt(2) into them to find the 2 slopes we need to compare

anonymous · Answer

$$-\sqrt{2} and \sqrt{2}$$

amistre64 · Answer

good.  now all we need to do is determine the angle between them

amistre64 · Answer

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