Group+D+Period+8

Welcome Group D Period 8! You must complete problems 4, 10, and 16 from section 4.6. Feel free to use the discussion forum to bounce ideas off one another. Use the space below to complete the assignment.

10) a) v=bwh v=xyz dv/dt=z(dx/dt(y)+(x)dy/dt) + xy(dz/dt) dv/dt=12 + 2(-8+3) __dv/dt=2m^3/s__ The rate at which the volume of this box changes when x=4, y=3, z=2, dx/dt= 1m/s, dy/dt=-2m/s, and dz/dt=1m/s is 2m^3/s.

b) A=2(xy)+2(xz)+2(yz) da/dt=2(dx/dt(y)+(x)dy/dt)+(dx/dt(z)+(x)dz/dt)+2(dy/dt(z)+(y)dz/dt) da/dt=[dx/dt(2y)+dy/dt(2x)]+[dx/dt(2z)+dz/dt(2x)]+[dy/dt(2z)+dz/dt(2y)] da/dt=6-16+4+8-8+6 __da/dt=0m^2/s__ The rate at which the surface area of the box changes when x=4, y=3, z=2, dx/dt= 1m/s, dy/dt=-2m/s, and dz/dt=1m/s is 0m^2/s.

c) s=(x^2+y^2+z^2)^.5 ds/dt=[.5(x^2+y^2+z^2)^-.5][2x(dx/dt)+2y(dy/dt)+2z(dz/dt)] ds/dt=(.5(29)^-.5)(8(1)+6(-2)+4(1) __ds/dt=0m/s__ The rate at which the diagonal is changing when x=4, y=3, z=2, dx/dt= 1m/s, dy/dt=-2m/s, and dz/dt=1m/s is 0m/s.

4. In part A you use implicit differentiation and product rule to write dP/dt in terms of dI/dt and dR/dt. In part B you re-write the problem so R is a function of P and I. Then you use implicit differentiation to write dR/dt in terms of dI/dt and dP/dt. At the end, both terms in the numerator have an I, so this can be cancelled with the denominator.