# 微積分新手遇到問題

1.Find the point where the hnction f (x, y ) = x^2y - 2xy - 3y + y^2 has the local minimum.

2.Consider a pyramid with height h and rectangular base with dimensions b and 2b . Find thevolume of the described solid.

3.Find the ranges of x where the function f ( x ) = x^4 - 2x^3 + x^2 - 2x + 1 is concave.

4..Compute the volume enclosed by the two cylinders: x2 + y2 = a2 and x2 + z2 = a2, a

### 1 個解答

• 最佳解答

1. First find its critical points where f_x=0 and f_y=0. i.e 2xy-2y=0 and x^2-2x-3+2y=0 => (x,y)=(3,0), (-1,0), or (1,2) ----3 critical points. Next, identify each critical point using @nd derivative test:

Preparing D(x,y)=(f_xx)(f_yy)-(f_xy)^2=4y-(2x-2)^2, evaluate D at critical points => D(3,0)=-1<0 so (3,0,f(3,0)) is a saddle point, not local extremal point; D(-1,0)-16<0, also a saddle point; D(1,2)=8>0, together with f_xx(1,2)=4>0, shows that (1,2,f(1,2))=(1,2,-4) is indeed a local minimum point on the graph of f.

2. A pyramid is after all a cone. So V=(area of the base)*(height)*(1/3)=2b^2/3.

3. I suppose you are asking " Find the ranges of x where the function f ( x ) = x^4 - 2x^3 + x^2 - 2x + 1 is concave up or down"?.

Using f"(x)=12x^2-12x+2, determine where it is positive or negative. Calcutating f"=0 we find x=(1/2)*(1+sqrt(1/3)) and x=(1/2)*(1-sqrt(1/3)). Therefore (i) f is concave upward on (-infinity,(1/2)*(1-sqrt(1/3))) and ((1/2)*(1+sqrt(1/3)), infinity); and (ii)f is concave downward on ((1/2)*(1-sqrt(1/3)), (1/2)*(1+sqrt(1/3))).

4. the volume enclosed by the two cylinders: x2 + y2 = a2 and x2 + z2 = a2 is the same as the volume enclosed by the two cylinders: x2 + y2 = a2 and y2 + z2 = a2.

I'm working on the later one just because it is easier for me to visualize.

The solid has symmetry with respect to x- y-, and z- axis. Therefore it suffices to find the volume of that solid in first octant and multiply the result by 8. Furthermore the part of the solid in the first octant is also symmetric with respect to y=x, because two cylinder intersect at a curve which is on plane y=z. So the desired volume V=16*{ the volume of the solid under the surface x2 + z2 = a2, [above the xy-plane, ] above the triangular region bounded by x=y, x=0, and x=a} = 16*int[x=0 to a] int[y=0 to x] { sqrt[a^2-x^2]} dydx. Since volume=double integral {z} dA, and z=sqrt[a^2-x^2].

The calculated result is V=16/3*a^3.