Higher number differential problem The solution leaks into a cylindrical barrel with a diameter of 10cm from a conical funnel with a depth of 18cm and a diameter of 12cm. At the beginning, the funnel is filled with solution. It is known that when the depth of the solution in the funnel is 12cm, the rate of surface decline is 1cm / min. it is not very grateful to ask what the rate of surface rise of the solution in the cylindrical barrel is at this time

Higher number differential problem The solution leaks into a cylindrical barrel with a diameter of 10cm from a conical funnel with a depth of 18cm and a diameter of 12cm. At the beginning, the funnel is filled with solution. It is known that when the depth of the solution in the funnel is 12cm, the rate of surface decline is 1cm / min. it is not very grateful to ask what the rate of surface rise of the solution in the cylindrical barrel is at this time

Let the depth of the liquid level in the conical funnel be h, the radius be r, and the liquid volume be V; The depth of the liquid level in the cylindrical barrel is h, the radius is r, and the liquid volume is v
v=1/3*pi*r^2*h
H / 18 = 2 * r / 12, i.e. H = 3 * r
v=1/27*pi*h^3
dv/dt=1/27*pi*3*h^2*dh/dt
V=pi*R^2*H
dV/dt=pi*R^2*dH/dt
Due to volume conservation
So DV / dt = DV / DT
1/27*pi*3*h^2*dh/dt=pi*R^2*dH/dt
1/9*h^2*dh/dt=R^2*dH/dt
dH/dt=1/9*h^2/R^2*dh/dt=12^2/3^2/5^2*1=0.64cm/min
Therefore, the rising rate of the solution surface in the cylindrical barrel is 0.64cm/min

High number differential Thanks ∫ (t-sint) ^ 2sintdt

∫（t-sint）^2sintdt =∫(t^2sint+sint^2sint-2tsint^2)dt=∫t^2sintdt+∫(1-cost^2)sintdt-2∫tsint^2dt=-∫t^2dcost-∫(1-cost^2)dcost-∫t*(1-cos2t)dt=-t^2cost+∫2costdt-cost+cost^3/3-t^2/2+∫tdsin2t=-t^2c...

The question of higher numbers ~ on the concept of differential Can you tell me the relationship between △ x △ y dy DX in popular words If I can find out where our teacher is, I won't ask questions. T. what's the relationship between Dy and △ x? The relationship between the four

The landlord can think like this: Δ X is (x1-x2), which is equivalent to taking a section on X with unlimited size, Δ Y similarly; In essence, DX takes a segment on X, but the length of this segment approaches 0, Dy is the same; So when Δ x, Δ When y approaches 0, Lim Δ x/ Δ Y = dy / DX, there is only this relationship between the four, that is, Dy and DX Δ X doesn't matter, DX and Δ It doesn't matter

Mathematical analysis of differential problems of multivariate functions What are the sufficient conditions for the differentiability of multivariate functions? You'd better have a detailed explanation

Partial derivative continuous → differentiable → existence of partial derivative

1.y=ln(1-x) 2.y=(e^x)sinx 3.y=(e^x)+sinx 4.y=lnsin(3x) 5.y=e^(-x)cos(3-x)

1)dy/dx=-1/(1-x)
2)dy/dx=e^x(sinx+cosx)
3)dy/dx=e^x+cosx
4)dy/dx=3cos(3x)/sin(3x)
5)dy/dx=-e^(-x)cos(3-x) +e^(-x)sin(3-x)

How to find the tangent vector equation of curve equation? How to find the normal vector equation of the surface equation?

If the tangent vector of the curve is given by the parametric equation, the variables can be derived from the parameters respectively. If it is given by the equation group, the implicit function of other variables on a variable can generally exist. Therefore, at this time, other variables can be regarded as the function of this variable, and the derivatives of other variables on the function of this variable can be derived from each equation of the equation group, Since other variables take this variable as a parameter, the tangent vector equation can be given by the method of parametric equation, and then the tangent vector can be obtained by bringing in the coordinates of the point
For the normal vector of the surface equation, just take the derivative of the equation on each variable, and then bring the coordinates of the point into the normal vector
What you said may be abstract. You just need to find a few examples combined with my understanding. I'm also reviewing these things, learning from each other and communicating with each other if you don't understand