Finding second derivative at (3,2)?
If x^2+y^2=25, what is d^2x/d^2y at point (3,2)?
I get -13/8, is this correct?
So sorry, I mixed up the point. The given point is (2,3) instead of (3,2).
If x^2+y^2=25, what is d^2x/d^2y at point (3,2)?
I get -13/8, is this correct?
So sorry, I mixed up the point. The given point is (2,3) instead of (3,2).
?
Favorite Answer
x^2 + y^2 = 25. 2x + 2yy' = 0, ie., x + yy' = 0...(1). 1 + yy'' +(y')^2 = 0...(2).
At (2,3), (1)--->2 + 3y' = 0, ie., y' = -(2/3) and (2)--->1 + 3y'' + (-2/3)^2 = 0, ie.,
y'' = -(1/3)(1 + 4/9) = -(1/3)(13/9) = - 13/27.
la console
x² + y² = 25 → when: x = 3
9 + y² = 25
y² = 16
y = ± 4
→ Point A (3 ; 4)
→ Point B (3 ; - 4)
x² + y² = 25 → when x is constant: → 2y.dy = 0
x² + y² = 25 → when y is constant: → 2x.dx = 0
2y.dy + 2x.dx = 0
2y.dy = - 2x.dx
dy/dx = - 2x/2y
dy/dx = - x/y
(dy/dx)² = - (x/y)²
d²y/d²x = - x²/y²
d²y/d²x = - x²/y² → @ the point A (3 ; 4)
d²y/d²x = - 9/16
d²y/d²x = - x²/y² → @ the point B (3 ; - 4)
d²y/d²x = - 9/16
alex
(3,2) is NOT on the curve !
Φ² = Φ+1
If (a,b) is a point on x²+y²=r², so a²+b²=r², then:
1) dy/dx at (a,b) is the slope of the tangent line at that point. The line has the equation ax+by=r², so dy/dx = -a/b and so dx/dy = -b/a.
2) d²y/dx² at (a,b) is the slope of the radial line at that point. The line has the equation bx-ay=0, so d²y/dx² = b/a, and so d²x/dy² = a/b.
Here d²x/dy² at (3,2) would equal 3/2 for x²+y²=13.
Pope
Your update gets us no closer to an answer. Neither (3, 2) nor (2, 3) satisfies the equation you gave. I am not saying that the answer cannot be found. What I mean is that the answer cannot even exist.