(1)
y的平方 =2+xy
2y(dy/dx) =x(dy/dx)+y(1)
(2y-x)(dy/dx)=y
dy/dx = y/(2y-x)
(2)
when dy/dx = 1/2
y/(2y-x) = 1/2
2y=2y-x
x=0
y的平方 =2+(0)y
y^2=2
y=(+/-)根2
(3)
Show when dy/dx = 0
no solution
(4)
consider the curve given by y的平方 =2+xy
consider the curve given by y的平方 =2+xy
(1) show that dy/dx = y/(2y-x)
(2) find all points (x,y) on the curve where the line tagent to the curve has slope 1/2
(3)show that there are no points (x,y) on the curve where the line tangent to the curve is horizontal
(4) let x and y be functions of time t that are related by the equation y的平方 =2+xy.at time t=5,the value of y is 3 and dy/dt=6.find the value of dx/dt at time t=5
(1) show that dy/dx = y/(2y-x)
(2) find all points (x,y) on the curve where the line tagent to the curve has slope 1/2
(3)show that there are no points (x,y) on the curve where the line tangent to the curve is horizontal
(4) let x and y be functions of time t that are related by the equation y的平方 =2+xy.at time t=5,the value of y is 3 and dy/dt=6.find the value of dx/dt at time t=5
英語(yǔ)人氣：245 ℃時(shí)間：2020-02-04 22:21:31
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