find the angle between the curves x²y=4,y(x²+4)=8
Answers
Step-by-step explanation:
1: Find intersection of those two curves.
The second equation can be rewritten as x^2 y + 4y = 8. This means that is this and x^2 y = 4 is satisfied simultaneously, 4y = 4, therefore y = 1. Since x^2 y = 4, x^2 = 4 and x is either 2 or -2. So we have two points of intersection, (2,1) and (-2,1). But it doesn’t really matter — both functions are even, so the angle would come out the same for either point. We’ll work with (2,1).
2: Find tangents of the curves in the point (2,1).
We need to first rewrite the equations as functions in y: y = 4/x^2 and y = 8/(x^2+4).
Then we must derive both. First is simple: y’ = -8/x^3. Second is harder, but we get y’ = -16x/(x^2+4)^2.
What we need is just the value of those derivations at x=2, and they are -1 for the first function and -1/2 for the second one.
We don’t even need the precise equation of the tangents. We can just move the intersection point to the origin and treat those lines as vectors. The first tangent corresponds to vector [1,-1] and the other to [1,-1/2], or [2,-1] to make things simpler.
When we take dot product of these vectors, we get 1*2 + (-1)*(-1) = 2 + 1 = 3. We know that dot product of vectors is equal to the product of their magnitudes and cosine of their angle; the magnitude of [1,-1] is sqrt(2) and magnitude of [2,-1] is sqrt(5), so we have:
3 = sqrt(2)*sqrt(5)*cos(angle)
3 = sqrt(10)*cos(angle)
cos(angle) = 3/sqrt(10)
That comes out to an angle of roughly 18.43 degrees. That is the angle in which those two curves intersect.
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Find intersection of those two curves.
The second equation can be rewritten as x^2 y + 4y = 8. This means that is this and x^2 y = 4 is satisfied simultaneously, 4y = 4, therefore y = 1. Since x^2 y = 4, x^2 = 4 and x is either 2 or -2. So we have two points of intersection, (2,1) and (-2,1). But it doesn’t really matter — both functions are even, so the angle would come out the same for either point. We’ll work with (2,1).
2: Find tangents of the curves in the point (2,1).
We need to first rewrite the equations as functions in y: y = 4/x^2 and y = 8/(x^2+4).
Then we must derive both. First is simple: y’ = -8/x^3. Second is harder, but we get y’ = -16x/(x^2+4)^2.
What we need is just the value of those derivations at x=2, and they are -1 for the first function and -1/2 for the second one.
We don’t even need the precise equation of the tangents. We can just move the intersection point to the origin and treat those lines as vectors. The first tangent corresponds to vector [1,-1] and the other to [1,-1/2], or [2,-1] to make things simpler.
When we take dot product of these vectors, we get 1*2 + (-1)*(-1) = 2 + 1 = 3. We know that dot product of vectors is equal to the product of their magnitudes and cosine of their angle; the magnitude of [1,-1] is sqrt(2) and magnitude of [2,-1] is sqrt(5), so we have:
3 = sqrt(2)*sqrt(5)*cos(angle)
3 = sqrt(10)*cos(angle)
cos(angle) = 3/sqrt(10)
That comes out to an angle of roughly 18.43 degrees. That is the angle in which those two curves intersect.