Given
r
(
t
)
=
−
4
t
2
i
+
e
−
2
t
j
+
4
t
k
and
u
(
t
)
=
2
t
3
i
+
e
−
9
t
j
+
3
t
3
k
, calculate
d
d
t
[
r
(
t
)
×
u
(
t
)
]
.

We are asked to calculate the derivative of the cross product of two vector-valued functions:

\[
r(t) = -4t^2 \hat{i} + e^{-2t} \hat{j} + 4t \hat{k}
\]
and
\[
u(t) = 2t^3 \hat{i} + e^{-9t} \hat{j} + 3t^3 \hat{k}.
\]

### Step 1: Cross Product of \( r(t) \) and \( u(t) \)
The cross product \( r(t) \times u(t) \) is given by the determinant of the following matrix:

\[
r(t) \times u(t) =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-4t^2 & e^{-2t} & 4t \\
2t^3 & e^{-9t} & 3t^3
\end{vmatrix}.
\]

The cross product is computed by expanding the determinant:

\[
r(t) \times u(t) = \hat{i} \begin{vmatrix} e^{-2t} & 4t \\ e^{-9t} & 3t^3 \end{vmatrix}
- \hat{j} \begin{vmatrix} -4t^2 & 4t \\ 2t^3 & 3t^3 \end{vmatrix}
+ \hat{k} \begin{vmatrix} -4t^2 & e^{-2t} \\ 2t^3 & e^{-9t} \end{vmatrix}.
\]

Now, calculate each of these 2x2 determinants.

### Step 2: Compute the 2x2 Determinants

1. For the \( \hat{i} \)-component:
\[
\begin{vmatrix} e^{-2t} & 4t \\ e^{-9t} & 3t^3 \end{vmatrix}
= e^{-2t}(3t^3) - e^{-9t}(4t) = 3t^3 e^{-2t} - 4t e^{-9t}.
\]

2. For the \( \hat{j} \)-component:
\[
\begin{vmatrix} -4t^2 & 4t \\ 2t^3 & 3t^3 \end{vmatrix}
= (-4t^2)(3t^3) - (4t)(2t^3) = -12t^5 - 8t^4 = -4t^4(3t + 2).
\]

3. For the \( \hat{k} \)-component:
\[
\begin{vmatrix} -4t^2 & e^{-2t} \\ 2t^3 & e^{-9t} \end{vmatrix}
= (-4t^2)(e^{-9t}) - (e^{-2t})(2t^3) = -4t^2 e^{-9t} - 2t^3 e^{-2t}.
\]

Thus, the cross product is:

\[
r(t) \times u(t) = \hat{i} \left( 3t^3 e^{-2t} - 4t e^{-9t} \right) - \hat{j} \left( -4t^4(3t + 2) \right) + \hat{k} \left( -4t^2 e^{-9t} - 2t^3 e^{-2t} \right).
\]

### Step 3: Derivative of the Cross Product

Now, we differentiate the result \( r(t) \times u(t) \) with respect to \( t \).

1. For the \( \hat{i} \)-component:
\[
\frac{d}{dt} \left( 3t^3 e^{-2t} - 4t e^{-9t} \right) = 9t^2 e^{-2t} + 3t^3(-2)e^{-2t} - 4e^{-9t} - 4t(-9)e^{-9t},
\]
which simplifies to:
\[
9t^2 e^{-2t} - 6t^3 e^{-2t} - 4e^{-9t} + 36t e^{-9t}.
\]

2. For the \( \hat{j} \)-component:
\[
\frac{d}{dt} \left( -4t^4(3t + 2) \right) = -4 \left( (3t + 2)(4t^3) + t^4(3) \right) = -16t^3(3t + 2) - 12t^4.
\]

3. For the \( \hat{k} \)-component:
\[
\frac{d}{dt} \left( -4t^2 e^{-9t} - 2t^3 e^{-2t} \right) = -8t e^{-9t} - 4t^2(-9)e^{-9t} - 6t^2 e^{-2t} - 2t^3(-2)e^{-2t},
\]
which simplifies to:
\[
-8t e^{-9t} + 36t^2 e^{-9t} - 6t^2 e^{-2t} + 4t^3 e^{-2t}.
\]

### Final Answer:

Thus, the derivative of the cross product is:

\[
\frac{d}{dt} \left( r(t) \times u(t) \right) = \hat{i} \left( 9t^2 e^{-2t} - 6t^3 e^{-2t} - 4e^{-9t} + 36t e^{-9t} \right) - \hat{j} \left( 16t^3(3t + 2) + 12t^4 \right) + \hat{k} \left( -8t e^{-9t} + 36t^2 e^{-9t} - 6t^2 e^{-2t} + 4t^3 e^{-2t} \right).
\]

Would you like to go over any specific part in detail?

Here are some questions to expand your understanding:
1. What is the general rule for differentiating a cross product of two vector functions?
2. How does the product rule apply to vector cross products?
3. What role does the exponential function play in vector calculus?
4. How do you simplify cross product expressions before differentiating?
5. How does each term in the cross product contribute to the final derivative?

**Tip**: When performing a cross product, always verify your vector components and determinant carefully to avoid sign errors.

Given r(t) = -4t^2 i + e^{-2t} j + 4t k and u(t) = 2t^3 i + e^{-9t} j + 3t^3 k, calculate d/dt [r(t) × u(t)].

Explore the step-by-step solution to calculating the derivative of the cross product of two vector functions, r(t) and u(t). This problem covers essential vector calculus concepts, including cross products, differentiation, and the application of the product rule. Suitable for students at the undergraduate level in Calculus III.

Math Problem Statement

Given r ( t )

− 4 t 2 i + e − 2 t j + 4 t k and u ( t )

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