assume x and y are functions of t. Evaluate dy/dt for 2xy--4x/3y^3=48, with conditions dx/dt=-6,x=3, y=-2
Just use the product rule and chain rule:
2xy - 4x/3y^3 = 48
2y dx/dt + 2x dy/dt - 4/3y^3 dx/dt + 4x/y^4 dy/dt = 0
Now just plug in the numbers and solve for dy/dt
d/dt (2xy-4x/3y^3)= d/dt (48)
2*d(xy)/dt - d(4x/3y^3)/dt = 0
(because derrvation of constant is 0)
2* ((dx/dt)*y + xdy/dt) -(((d(4x)/dt)*3y^3^-4x*d(3y^3)/dt))/(3y^3)^2)=0
2*((-6)*(-2)+3*(dy/dt))-(4*(dx/dt)*(3y^3)-4*x*3*(3*y^2)*(dy/dt) )/9y^6 =0
2*(12+3*(dy/dt))-(4*(-6)*(3*(-2)^3)-4*3*3*(3*(-2)^2)*(dy/dt) )/(9*(-2)^6) =0
2*(12+3*(dy/dt))-(4*(-6)*(3*(-2)^3)-4*3*3*(3*(-2)^2)*(dy/dt) )/(9*(-2)^6) =0
24+6*(dy/dt)-((-24)*(-24)-36*(12)*(dy/dt) )/(9*64) =0
24+6*(dy/dt)-(576-432*(dy/dt) )/(576) =0
24+6*(dy/dt)-(576/576-432*(dy/dt)/576) =0
24+6*(dy/dt)-(1-3*(dy/dt)/4) =0
24+6*(dy/dt)-(1-3*(dy/dt)/4) =0 =0
23+(6+(3/4))*(dy/dt)=0
23+(24/4 + 3/4) *(dy/dt)=0
(27/4 ) *(dy/dt)= -23
(dy/dt)= (-23 ) /(27/4)
(dy/dt) = (-23/1 ) /(27/4)
(dy/dt) = (-23*4/27 )
(dy/dt) = (-92/27
du/dx=-3.41 roundet on 2-nd decimal place )
That's a lot of places to make mistakes. I get
2y dx/dt + 2x dy/dt - 4/3y^3 dx/dt + 4x/y^4 dy/dt
2y(1 - 2/3y^4) dx/dt + 2x(1+2/y^4) dy/dt = 0
-4(23/24)(-6) + 6(9/8) dy/dt = 0
dy/dt = 27/92
Maybe you can work out whether we are both wrong!
Actually, I plugged in the line
2*(12+3*(dy/dt))-(4*(-6)*(3*(-2)^3)-4*3*3*(3*(-2)^2)*(dy/dt) )/(9*(-2)^6) = 0
at wolframalpha and got dy/dx = -92/27
So, maybe you better do it yourself... Watch out for details.
To evaluate dy/dt for the given equation 2xy - 4x/3y^3 = 48, we can use implicit differentiation.
Step 1: Rewrite the equation in terms of y and its derivative dy/dt:
2xy - 4x/3y^3 = 48
Multiply both sides by y^3:
2xy * y^3 - (4x/3y^3) * y^3 = 48 * y^3
2xy^4 - 4x = 48y^3
Step 2: Differentiate both sides of the equation with respect to t, treating y as a function of t:
d/dt(2xy^4 - 4x) = d/dt(48y^3)
To differentiate the left side, we will use the product rule:
d/dt(2xy^4 - 4x) = d/dt(2xy^4) - d/dt(4x)
The derivative of 2xy^4 with respect to t involves using both the product rule and the chain rule:
d/dt(2xy^4) = 2x * d/dt(y^4) + y^4 * d/dt(x)
Step 3: Calculate the derivatives using the given conditions dx/dt = -6, x = 3, and y = -2:
The derivative of y^4 with respect to t is found using the chain rule:
d/dt(y^4) = 4y^3 * (dy/dt)
Substituting the given values:
d/dt(2xy^4) = 2(3) * 4(-2)^3 * (dy/dt) = -384(dy/dt)
The derivative of x with respect to t is given as dx/dt = -6.
d/dt(x) = -6
Substituting all these derivatives back into our equation:
-384(dy/dt) - 4(-6) = 48(3)^3
-384(dy/dt) + 24 = 48(27)
-384(dy/dt) + 24 = 1296
-384(dy/dt) = 1296 - 24
-384(dy/dt) = 1272
Dividing both sides by -384:
dy/dt = 1272 / -384
dy/dt = -3.3125
Therefore, the value of dy/dt is approximately -3.3125.