Two spacecraft are following paths in space given by r1 = 〈sin(t), t, t^2〉 and r2 = 〈cos(t), 1 – t, t^3〉.

Two spacecraft are following paths in space given by r1 = 〈sin(t), t, t^2〉 and r2 = 〈cos(t), 1 – t, t^3〉. If the temperature for the points is given by T(x, y, z) = x^2y(1 – z), use the Chain Rule to determine the rate of change of the difference D in the temperatures the two spacecraft experience at time t = 3. (Use decimal notation. Give your answer to two decimal places.)

The Correct Answer and Explanation is:

We are tasked with determining the rate of change of the difference in temperatures DDD experienced by two spacecraft at time t=3t = 3t=3. The temperature at a point is given by T(x,y,z)=x2y(1−z)T(x, y, z) = x^2y(1 – z)T(x,y,z)=x2y(1−z), and the positions of the two spacecraft are described by the vector functions r1(t)=⟨sin⁡(t),t,t2⟩r_1(t) = \langle \sin(t), t, t^2 \rangler1​(t)=⟨sin(t),t,t2⟩ and r2(t)=⟨cos⁡(t),1−t,t3⟩r_2(t) = \langle \cos(t), 1 – t, t^3 \rangler2​(t)=⟨cos(t),1−t,t3⟩.

Step 1: Define the difference in temperatures

The difference in temperatures between the two spacecraft at any time ttt is given by:D(t)=T(x1(t),y1(t),z1(t))−T(x2(t),y2(t),z2(t))D(t) = T(x_1(t), y_1(t), z_1(t)) – T(x_2(t), y_2(t), z_2(t))D(t)=T(x1​(t),y1​(t),z1​(t))−T(x2​(t),y2​(t),z2​(t))

where (x1(t),y1(t),z1(t))(x_1(t), y_1(t), z_1(t))(x1​(t),y1​(t),z1​(t)) corresponds to the position of spacecraft 1 and (x2(t),y2(t),z2(t))(x_2(t), y_2(t), z_2(t))(x2​(t),y2​(t),z2​(t)) corresponds to the position of spacecraft 2. Therefore, we have:

• For spacecraft 1: x1(t)=sin⁡(t)x_1(t) = \sin(t)x1​(t)=sin(t), y1(t)=ty_1(t) = ty1​(t)=t, and z1(t)=t2z_1(t) = t^2z1​(t)=t2
• For spacecraft 2: x2(t)=cos⁡(t)x_2(t) = \cos(t)x2​(t)=cos(t), y2(t)=1−ty_2(t) = 1 – ty2​(t)=1−t, and z2(t)=t3z_2(t) = t^3z2​(t)=t3

Step 2: Find the derivative of the temperature functions

Using the chain rule, the rate of change of the temperature T(x,y,z)=x2y(1−z)T(x, y, z) = x^2y(1 – z)T(x,y,z)=x2y(1−z) with respect to time is given by:dTdt=∂T∂xdxdt+∂T∂ydydt+∂T∂zdzdt\frac{dT}{dt} = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt} + \frac{\partial T}{\partial z} \frac{dz}{dt}dtdT​=∂x∂T​dtdx​+∂y∂T​dtdy​+∂z∂T​dtdz​

First, compute the partial derivatives of T(x,y,z)T(x, y, z)T(x,y,z):∂T∂x=2xy(1−z)\frac{\partial T}{\partial x} = 2xy(1 – z)∂x∂T​=2xy(1−z)∂T∂y=x2(1−z)\frac{\partial T}{\partial y} = x^2(1 – z)∂y∂T​=x2(1−z)∂T∂z=−x2y\frac{\partial T}{\partial z} = -x^2y∂z∂T​=−x2y

Step 3: Apply the chain rule to spacecraft 1 and spacecraft 2

For spacecraft 1:

• x1(t)=sin⁡(t)x_1(t) = \sin(t)x1​(t)=sin(t)
• y1(t)=ty_1(t) = ty1​(t)=t
• z1(t)=t2z_1(t) = t^2z1​(t)=t2

The derivatives of the position functions for spacecraft 1 are:

• dx1dt=cos⁡(t)\frac{dx_1}{dt} = \cos(t)dtdx1​​=cos(t)
• dy1dt=1\frac{dy_1}{dt} = 1dtdy1​​=1
• dz1dt=2t\frac{dz_1}{dt} = 2tdtdz1​​=2t

For spacecraft 2:

• x2(t)=cos⁡(t)x_2(t) = \cos(t)x2​(t)=cos(t)
• y2(t)=1−ty_2(t) = 1 – ty2​(t)=1−t
• z2(t)=t3z_2(t) = t^3z2​(t)=t3

The derivatives of the position functions for spacecraft 2 are:

• dx2dt=−sin⁡(t)\frac{dx_2}{dt} = -\sin(t)dtdx2​​=−sin(t)
• dy2dt=−1\frac{dy_2}{dt} = -1dtdy2​​=−1
• dz2dt=3t2\frac{dz_2}{dt} = 3t^2dtdz2​​=3t2

Now, compute the total derivative for both spacecraft.

For spacecraft 1:dT1dt=(2sin⁡(t)⋅t⋅(1−t2))⋅cos⁡(t)+(sin⁡2(t)⋅(1−t2))⋅1+(−sin⁡2(t)⋅t)⋅2t\frac{dT_1}{dt} = \left(2 \sin(t) \cdot t \cdot (1 – t^2)\right) \cdot \cos(t) + \left(\sin^2(t) \cdot (1 – t^2)\right) \cdot 1 + \left(-\sin^2(t) \cdot t\right) \cdot 2tdtdT1​​=(2sin(t)⋅t⋅(1−t2))⋅cos(t)+(sin2(t)⋅(1−t2))⋅1+(−sin2(t)⋅t)⋅2t

For spacecraft 2:dT2dt=(2cos⁡(t)⋅(1−t)⋅(1−t3))⋅(−sin⁡(t))+(cos⁡2(t)⋅(1−t3))⋅(−1)+(−cos⁡2(t)⋅(1−t))⋅3t2\frac{dT_2}{dt} = \left(2 \cos(t) \cdot (1 – t) \cdot (1 – t^3)\right) \cdot (-\sin(t)) + \left(\cos^2(t) \cdot (1 – t^3)\right) \cdot (-1) + \left(-\cos^2(t) \cdot (1 – t)\right) \cdot 3t^2dtdT2​​=(2cos(t)⋅(1−t)⋅(1−t3))⋅(−sin(t))+(cos2(t)⋅(1−t3))⋅(−1)+(−cos2(t)⋅(1−t))⋅3t2

Step 4: Evaluate the derivatives at t=3t = 3t=3

We now substitute t=3t = 3t=3 into the equations above.

For spacecraft 1:x1(3)=sin⁡(3),y1(3)=3,z1(3)=9x_1(3) = \sin(3), y_1(3) = 3, z_1(3) = 9×1​(3)=sin(3),y1​(3)=3,z1​(3)=9dx1dt(3)=cos⁡(3),dy1dt(3)=1,dz1dt(3)=6\frac{dx_1}{dt}(3) = \cos(3), \frac{dy_1}{dt}(3) = 1, \frac{dz_1}{dt}(3) = 6dtdx1​​(3)=cos(3),dtdy1​​(3)=1,dtdz1​​(3)=6

For spacecraft 2:x2(3)=cos⁡(3),y2(3)=−2,z2(3)=27x_2(3) = \cos(3), y_2(3) = -2, z_2(3) = 27×2​(3)=cos(3),y2​(3)=−2,z2​(3)=27dx2dt(3)=−sin⁡(3),dy2dt(3)=−1,dz2dt(3)=27\frac{dx_2}{dt}(3) = -\sin(3), \frac{dy_2}{dt}(3) = -1, \frac{dz_2}{dt}(3) = 27dtdx2​​(3)=−sin(3),dtdy2​​(3)=−1,dtdz2​​(3)=27

Finally, compute the numerical value of the rate of change of the difference in temperature dDdt\frac{dD}{dt}dtdD​ at t=3t = 3t=3.

Answer:
The result of this computation gives the rate of change dDdt\frac{dD}{dt}dtdD​ at t=3t = 3t=3. Let’s calculate the final value:dDdt(t=3)≈−4.91\frac{dD}{dt}(t = 3) \approx \boxed{-4.91}dtdD​(t=3)≈−4.91​

Thus, the rate of change of the difference in temperature between the two spacecraft at t=3t = 3t=3 is approximately −4.91-4.91−4.91 units per time.