Math  /  Calculus

QuestionVerify the product law for differentiation, (AB)=AB+AB(\mathbf{A B})^{\prime}=\mathbf{A}^{\prime} \mathbf{B}+\mathbf{A} \mathbf{B}^{\prime} where A(t)=[3t3t1t31t]\mathbf{A}(t)=\left[\begin{array}{rr}3 t & 3 t-1 \\ t^{3} & \frac{1}{t}\end{array}\right] and B(t)=[1t1+t2t23t3]\mathbf{B}(t)=\left[\begin{array}{rr}1-t & 1+t \\ 2 t^{2} & 3 t^{3}\end{array}\right].
To calculate (AB)(\mathbf{A B})^{\prime}, first calculate AB\mathbf{A B}. AB=A B= \square Now take the derivative of ABA B to find (AB)(A B)^{\prime}. (AB)=(A B)^{\prime}= \square To calculate AB+AB\mathbf{A}^{\prime} \mathbf{B}+\mathbf{A B ^ { \prime }}, first calculate A\mathbf{A}^{\prime}. A=A^{\prime}= \square Now find AB\mathbf{A}^{\prime} \mathbf{B}. AB=A^{\prime} B= \square Now find B\mathbf{B}^{\prime}. B=\mathbf{B}^{\prime}=\square \square Now calculate AB\mathbf{A B}^{\prime}. AB=\mathbf{A B}^{\prime}=\square \square Finally, find AB+ABB\mathbf{A}^{\prime} \mathbf{B}+\mathbf{A B} \mathbf{B}^{\prime}. AB+AB=A^{\prime} B+A B^{\prime}= \square

Studdy Solution

STEP 1

1. We are given two matrix functions A(t)\mathbf{A}(t) and B(t)\mathbf{B}(t).
2. We need to verify the product rule for differentiation for these matrix functions.
3. Differentiation and matrix multiplication are performed element-wise.

STEP 2

1. Calculate the product AB\mathbf{A B}.
2. Differentiate the product (AB)(\mathbf{A B})^{\prime}.
3. Calculate the derivative of A\mathbf{A}, A\mathbf{A}^{\prime}.
4. Calculate the product AB\mathbf{A}^{\prime} \mathbf{B}.
5. Calculate the derivative of B\mathbf{B}, B\mathbf{B}^{\prime}.
6. Calculate the product AB\mathbf{A B}^{\prime}.
7. Verify the product law by calculating AB+AB\mathbf{A}^{\prime} \mathbf{B} + \mathbf{A B}^{\prime}.

STEP 3

Calculate the product AB\mathbf{A B}:
A=[3t3t1t31t],B=[1t1+t2t23t3]\mathbf{A} = \begin{bmatrix} 3t & 3t-1 \\ t^3 & \frac{1}{t} \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 1-t & 1+t \\ 2t^2 & 3t^3 \end{bmatrix}
AB=[(3t)(1t)+(3t1)(2t2)(3t)(1+t)+(3t1)(3t3)(t3)(1t)+(1t)(2t2)(t3)(1+t)+(1t)(3t3)]\mathbf{A B} = \begin{bmatrix} (3t)(1-t) + (3t-1)(2t^2) & (3t)(1+t) + (3t-1)(3t^3) \\ (t^3)(1-t) + \left(\frac{1}{t}\right)(2t^2) & (t^3)(1+t) + \left(\frac{1}{t}\right)(3t^3) \end{bmatrix}
Simplify each element:
AB=[3t3t2+6t32t23t+3t2+9t43t3t3t4+2tt3+t4+3t2]\mathbf{A B} = \begin{bmatrix} 3t - 3t^2 + 6t^3 - 2t^2 & 3t + 3t^2 + 9t^4 - 3t^3 \\ t^3 - t^4 + 2t & t^3 + t^4 + 3t^2 \end{bmatrix}

STEP 4

Differentiate AB\mathbf{A B} with respect to tt:
(AB)=[(36t+18t24t)(3+6t+36t39t2)(3t24t3+2)(3t2+4t3+6t)](\mathbf{A B})^{\prime} = \begin{bmatrix} (3 - 6t + 18t^2 - 4t) & (3 + 6t + 36t^3 - 9t^2) \\ (3t^2 - 4t^3 + 2) & (3t^2 + 4t^3 + 6t) \end{bmatrix}
Simplify:
(AB)=[310t+18t23+6t+36t39t23t24t3+23t2+4t3+6t](\mathbf{A B})^{\prime} = \begin{bmatrix} 3 - 10t + 18t^2 & 3 + 6t + 36t^3 - 9t^2 \\ 3t^2 - 4t^3 + 2 & 3t^2 + 4t^3 + 6t \end{bmatrix}

STEP 5

Calculate the derivative of A\mathbf{A}, A\mathbf{A}^{\prime}:
A=[333t21t2]\mathbf{A}^{\prime} = \begin{bmatrix} 3 & 3 \\ 3t^2 & -\frac{1}{t^2} \end{bmatrix}

STEP 6

Calculate the product AB\mathbf{A}^{\prime} \mathbf{B}:
AB=[333t21t2][1t1+t2t23t3]\mathbf{A}^{\prime} \mathbf{B} = \begin{bmatrix} 3 & 3 \\ 3t^2 & -\frac{1}{t^2} \end{bmatrix} \begin{bmatrix} 1-t & 1+t \\ 2t^2 & 3t^3 \end{bmatrix}
=[3(1t)+3(2t2)3(1+t)+3(3t3)3t2(1t)1t2(2t2)3t2(1+t)1t2(3t3)]= \begin{bmatrix} 3(1-t) + 3(2t^2) & 3(1+t) + 3(3t^3) \\ 3t^2(1-t) - \frac{1}{t^2}(2t^2) & 3t^2(1+t) - \frac{1}{t^2}(3t^3) \end{bmatrix}
Simplify:
=[33t+6t23+3t+9t33t23t323t2+3t33t]= \begin{bmatrix} 3 - 3t + 6t^2 & 3 + 3t + 9t^3 \\ 3t^2 - 3t^3 - 2 & 3t^2 + 3t^3 - 3t \end{bmatrix}

STEP 7

Calculate the derivative of B\mathbf{B}, B\mathbf{B}^{\prime}:
B=[114t9t2]\mathbf{B}^{\prime} = \begin{bmatrix} -1 & 1 \\ 4t & 9t^2 \end{bmatrix}

STEP 8

Calculate the product AB\mathbf{A B}^{\prime}:
AB=[3t3t1t31t][114t9t2]\mathbf{A B}^{\prime} = \begin{bmatrix} 3t & 3t-1 \\ t^3 & \frac{1}{t} \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 4t & 9t^2 \end{bmatrix}
=[3t(1)+(3t1)(4t)3t(1)+(3t1)(9t2)t3(1)+(1t)(4t)t3(1)+(1t)(9t2)]= \begin{bmatrix} 3t(-1) + (3t-1)(4t) & 3t(1) + (3t-1)(9t^2) \\ t^3(-1) + \left(\frac{1}{t}\right)(4t) & t^3(1) + \left(\frac{1}{t}\right)(9t^2) \end{bmatrix}
Simplify:
=[3t+12t24t3t+27t39t2t3+4t3+9t]= \begin{bmatrix} -3t + 12t^2 - 4t & 3t + 27t^3 - 9t^2 \\ -t^3 + 4 & t^3 + 9t \end{bmatrix}

STEP 9

Verify the product law by calculating AB+AB\mathbf{A}^{\prime} \mathbf{B} + \mathbf{A B}^{\prime}:
AB+AB=[33t+6t23+3t+9t33t23t323t2+3t33t]+[3t+12t24t3t+27t39t2t3+4t3+9t]\mathbf{A}^{\prime} \mathbf{B} + \mathbf{A B}^{\prime} = \begin{bmatrix} 3 - 3t + 6t^2 & 3 + 3t + 9t^3 \\ 3t^2 - 3t^3 - 2 & 3t^2 + 3t^3 - 3t \end{bmatrix} + \begin{bmatrix} -3t + 12t^2 - 4t & 3t + 27t^3 - 9t^2 \\ -t^3 + 4 & t^3 + 9t \end{bmatrix}
=[310t+18t23+6t+36t39t23t24t3+23t2+4t3+6t]= \begin{bmatrix} 3 - 10t + 18t^2 & 3 + 6t + 36t^3 - 9t^2 \\ 3t^2 - 4t^3 + 2 & 3t^2 + 4t^3 + 6t \end{bmatrix}
The result matches (AB)(\mathbf{A B})^{\prime}, confirming the product law.

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