𝑥 and 𝑦 are differentiable functions of 𝑡.

Answer 1

Let differentiate by t given equation: d/dt(6y²) = d/dt(x⁵ - x⁷); 12ydy/dt = (5x⁴ - 7x⁶)dxdt.

Let substitude given values: x = 1/5 in given equation to get y > 0: 6y² = (1/5)⁵ - (1/5)⁷,

y² = 5120 10^-8; y = 32√5·10^-4

So dy/dt = (5·0.2⁴ - 7·0.2⁶)·6/( 32√5·10^-4) = 354√5/125 = 6.333

Answer 2

We will first find dy/dx, then calculate dy/dt = dy/dx dx/dt

We can find dy/dx either by first writing y as a function of x or using implicit differentiation.

Let's do it both ways and show that the results are the same.

Writing y as a function of x, as 6y² = x⁵ - x⁷ and y > 0,

y = √(( x⁵ - x⁷ )/6)

Then, dy/dx = (5x⁴ - 7x⁶ )/(12√(( x⁵ - x⁷ )/6)). Note that this equals (5x⁴ - 7x⁶ )/(12 y)

If we use implicit differentiation, we get 12y dy = 5x⁴ - 7x⁶ or

dy/dx = (5x⁴ - 7x⁶ )/(12y)

This equals our result calculated the other way.

We are given that x = 1/5 and dx/dt = 6

When x = 1/5, y = √(( x⁵ - x⁷ )/6) = √(( 1/5⁵ - 1/5⁷ )/6) = 5^-7/2 * √((25-1)/6) = 2 * 5^-7/2

Then, dy/dx = (5x⁴ - 7x⁶ )/(12y) = 5^-6 (5*25 - 7)/(12 * 2 * 5^-7/2 ) = 5^-5/2 (59/12)

dx/dt = 6

Thus, dy/dt = dy/dx * dx/dt = 5^-5/2 (59/12) * 6 = 59 * 5^-5/2 /2 or 59/(2*5^5/2) or 59√5/250

Answer 3

Using 6y² = x⁵ - x⁷ and plugging in x = 1/5 we get:

6y² = (1/5)⁵ - (1/5)⁷

6y² = 1/3125 - 1/78125

6y² = 25/78125 - 1/78125

6y² = 24/78125

y² = 4/78125

y = 2/(125√5)

Now, take the derivative using implicit differentiation as follows:

So dy/dt = [5(1/5)⁴•(6) - 7(1/5)⁶•(6)]/[12•2/(125√5)] = (59√5)/250