Group+D+Period+6

Welcome Group D Period 6! 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.

4. **The power P (watts) of an electric circuit is related to the circuit's resistance R (ohms) and current I (amperes) by the equation P=R(I^2)**

a) **How is dP/dt related to dR/dt and dI/dt?**

In order to see how dP/dt relates to dR/dt and dI/dt we must take the derivative of the original euqauation. The derivative of P=RI^2 is equal to

dP/dt = (dR/dt)(I^2) + IR(dI/dt)

b) **How is dR/dt related to dI/dt if P is constant?**

In order to find how dR/dt relates to dI/dt we must set the equation equal to R which changes it to R = P(I^-2). Now we take the derivative of that function which looks like this.

dR/dt = (-2P/I^3)(dI/dt)

(We don't take the derivative of P because in this problem P is a constant)

10. x= 4m y= 3m z = 2m; dx/dt = 1m/s dy/dt = -2m/s dz/dt = 1m/s

a) V = x*y*z dv/dt = (dx/dt)(yz) + (dy/dt)(xz) + (dz/dt)(xy) = (1m/s)(3m)(2m) + (-2m/s)(4m)(2m) + (1m/s)(4m)(3m)

(6m^3/s) - (16m^3/s) + (12m^3/s)

 * = 2m^3/s**

b) As = 2xy + 2yz + 2xz dAs/dt = ((2x)(dy/dt) + (2dx/dt)(y)) + ((2y)(dz/dt) + (2dy/dt)(z)) + ((2z)(dx/dt) + (2dz/dt)(x)) = ((2)(4m)(-2m/s) + (2)(1m/s)(3m)) + ((2)(3m)(1m/s) + (2)(-2m/s)(2m)) + ((2)(2m)(1m/s) + (2)(1m/s)(4m)) = (-16m^2/s) + (6m^2/s) + (6m^2/s) + (-8m^2/s) + (4m^2/s) + (8m^2/s) = **0m^2/s**

c) s = (x^2 + y^2 + z^2)^(1/2) ds/dt = (1/2)(x^2 + y^2 + z^2)^(-1/2)*((2x)(dx/dt) + (2y)(dy/dt) + (2z)(dz/dt)) = ((x)(dx/dt) + (y)(dy/dt) + (z)(dz/dt))*(x^2 + y^2 + z^2)^(-1/2) = ((4m)(1m/s) + (3m)(-2m/s) + (2m)(1m/s))*((4m)^2 + (3m)^2 + (2m)^2)^(-1/2) = ((4m^2/s) + (-6m^2/s) + (2m^2/s)) * (16m^2 + 9m^2 + 4m^2)^(-1/2) = (0m^2/s)/ (29m^2)(1/2) = (0m^2/s)/ 5.38m = **0m/s**


 * 16. Sand falls from a conveyor belt at the rate of 10 m^3/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Give your answer in cm/min.**

a) Since it is a conical pile, we know that V=1/3(Pi)r^3h We are looking for dr/dt, so we need to take the derivative of this, which is
 * dV/dt=(Pi)r^2h(dr/dt)+1/3(pi)r^3(dh/dt)

We also know that the piles height = 3/8d, or h=3/4r This means that when the pile is 4 m high, we can evaluate for r by: We plug this into our first equation to find: We take the derivative of this: Plug 10 m^3/min into dV/dt and solve for dr/dt: We plug in 16/3 for r as we found earlier: We multiply this by 100 to change it to cm: b) Now to find dh/dt, we simply plug our dr/dt into our equation earlier of h=3/4r by taking the derivative: Plug in 1500/32(pi) for dr/dt:
 * 4=3/4r, which is equivalent to 16/3=r
 * V=1/3(pi)r^3(3/4r)=1/4(pi)r^3
 * dV/dt=3/4(pi)r^2(dr/dt)
 * 10=3/4(pi)r^2(dr/dt)
 * 40/3(pi)=r^2dr/dt
 * 40/3(pi)=(16/3)^2(dr/dt)
 * 40/3(pi)=256/9 dr/dt
 * dr/dt= 15/32(pi) m^3/min
 * dr/dt= 1500/32(pi) cm^3/min**
 * dh/dt=3/4(dr/dt)
 * dh/dt= 3/4(1500/32(pi))=11.19cm^3/min**