= (16) / (11) , - (16) / (5) ... for values of y = 6, -2,
since x = 4 in x²+xy-y² = 4 gives
16+4y-y² = 4,
y² -4y -16 +4= 0,
(y -2)² = 4² ,
y -2 = ±4,
y = 6, -2.
-
Sqdancefan say: You assume that both x and y are functions of t, and you use the chain rule to differentiate. Then substitute the given values and evaluate.
1) dy/dt = (4x +8)*dx/dt
.. at x=6, dx/dt=13, this is
.. dy/dt = (4*6 +8)*13 = 416
2) dy/dt = 1/(2√x)*dx/dt
.. at x=9, dx/dt=2, this is
.. dy/dt = 1/(2√9)*2 = 1/3
3) y^2 -xy +(4 -x^2) = 0
.. y = (x ±√(x^2 -4(4 -x^2)))/2 . . . it seems easiest to solve for y first
.. dy/dt = (1 ±5x/√(5x^2 -16))/2*dx/dt . . . differentiate
.. at x=4, dx/dt=2, this is
.. dy/dt = (1 ±5*4/√(5*4^2 -16))/2*2 = 1 ±2.5 = {-3/2, 7/2}
-
Alvin say: Updated: corrected 3 mistakes
First two are simple uses of the chain rule
1) when y=2x^2+8x+4 and dx/dt=13 when x=6
dy/dx = 4x + 8
dx/dt = 13
at x = 6
dy/dx= 4*6+ 8 = 32
dx/dt = 13
Using chain rule
answer:
dy/dx * dx/dt = dy/dt = 32*13 = 416
2) where y=√(x) -3 and dx/dt=2 when x=9
y = (√(x) ) - 3 = x^(1/2) - 3
dy/dx = (1/2) x^(-1/2)
at x =9 dy/dx = (1/2)* (1/ sqrt(9) ) = 1/2* (1/3) = 1/6
dx/dt = 2
answer:
chain rule dy/dx * dx/dt= dy/dt = 2* 1/6 = 1/3
3) where x^2+xy-y^2=4 and dx/dt=2 when x=4
Use implicit derivative to find dy/dx
2x + x*dy/dx + y - 2y*dy/dx = 0
move dy/dx term to one side
-2x-y = x*dy/dx -2y*dy/dx = (x-2y) dy/dx
dy/dx = (-2x-y) / (x -2y)
at x = 4
x^2+xy-y^2=4
16 + 4y -y^2 = 4
y^2 -4y -16 + 4 = 0
y^2 -4y -12 =0
(y -6)*(y+ 2) = 0
y = 6 or y =-2 when x =4
so at (4,6)
dy/dx = (-2*4 -6) / (4 -2*6) = (-14/ -8) = 7/4
so at (4, -2)
dy/dx = (-2*4 - (-2))/ (4 -2*(-2)) = (-6/ 8) = -3/4
so at(4,6) , dy/dt = (2* 7/4) = 7/2
so at (4, -2) dy/dt = (2*-3/4) = -3/2
you have two answers 7/2 at (4,6) and -3/2 at (4,-2)
Check my work
-
alex say: 1/
dy/dt = (dy/dx)*(dx/dt) = (4x+8)*(13) = (4*6+8)*13 =...
2/ similar 1/
3/
2x(dx/dt)+x(dy/dt)+y(dx/dt)-2y(dy/dt)=0
...you can continue
-