$(5)(See$ Figure $1)$ Suppose $(t)is$ traversed with constant speed. Let $v(t)be$ the velocity

$of(t)at$ time $t.Ifp=(t_1)=(t_2)$ for $v(t_1)=v(t_2)(t)\frac{dv}{dt}(t_1)=\frac{dv}{dt}(t_2)f(x,y,z)f=0R^{3}g(x,y,z)g=0g(x,y,z)=2f(x,y,z)g(x,y,z)=f(2x,2y,2z)g(x,y,z)=f(x^2,y^2,z^2)t=0(t)R^3($:$1,-1,1$:$)(0,1,0)f(x,y,z)f((t))t=0f=x+y+z^{2}f=x+y^2+zf=x^2+y+z{t}_1,$ then $it$ must $be$ true that:

$(A)v(t_1)=v(t_2)$

$(B)$ The acceleration $of(t)is$ always zero

$(C)\frac{dv}{dt}(t_1)=\frac{dv}{dt}(t_2)$

$(D)$ Two $of(A)-(C)$

$(E)$ None $of(A)-(D)$

Reason:

$(6)$ Suppose $f(x,y,z)$ satisfies $f=0$ everywhere $on{R}^3.$ The following $g(x,y,z)$ also

satisfies $g=0$ everywhere:

$(A)g(x,y,z)=2f(x,y,z)$

$(B)g(x,y,z)=f(2x,2y,2z)$

$(C)g(x,y,z)=f(x^2,y^2,z^2)$

$(D)$ Two $of(A)-(C).$

$(E)$ All $of$ the above.

Reason:

$(7)Att=0,$ a curve $(t)in{R}^3$ has velocity $($:$1,-1,1$:$)at$ point $(0,1,0).$ For which functions

$f(x,y,z)$ does $f((t))$ have derivative equal $to0att=0?$

$(A)f=x+y+z^2$

$(B)f=x+y^2+z$

$(C)f=x^2+y+z$

$(D)$ Two $of(A)-(C)$

$(E)$ None $of(A)-(C)$

Reasor$*$