已知“syms t; x=cos(t); y=sin(t); z=t; xt=diff(x,'t'); yt=diff(y,'t'); zt=diff(z,'t'); f=z^2/(x^2+y^2); g=sqrt(xt^2+yt^2+zt^2); I=int(f*g,t,0,2*pi)”,则下列说法正确的是【】
举一反三
- 已知“syms x y z t a b; x=a*cos(t); y=a*sin(t); z=3*t; dx=diff(x,'t'); dy=diff(y,'t'); dz=diff(z,'t'); f=y*dx-x*dy+(x+y+z)*dz; t1=0; t2=2*pi; W=int(f,t,t1,t2)”,则正确的说法是【】
- 实验命令“xt=@(t)sin(2*t); yt=@(t)cos(2*t); zt=@(t)t; fplot3(xt,yt,zt,[0 20pi])”,所绘制的图形是【 】
- 曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- 设\(z = f(x,y)\)，\(x = \sin t\)，\(y = {t^3}\)，则全导数\( { { dz} \over {dt}} = \) A: \({f'_x} \sin t+ 3{t^2}{f'_y}\) B: \({f'_x} \cos t+ {t^2}{f'_y}\) C: \({f'_x} \cos t+ 3{t^2}{f'_y}\) D: \({f'_y} \cos t+ 3{t^2}{f'_x}\)
- 已知函数[img=102x27]18030256dad01f2.png[/img]，求其三阶导数，下面命令正确的是（） A: syms t; G=simplify(diff(t^2*sin(t),t,3)) B: syms t; G=simplify(int(t^2*sin(t),t,3)) C: syms t; G=simplify(diff(t^2*sin(t),t)) D: syms t; G=simplify(int(t^2*sin(t),t))