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Chapter 11: Problem 40

Compute the following derivatives. $$\frac{d}{d t}\left(\left(t^{3} \mathbf{i}+6 \mathbf{j}-2 \sqrt{t} \mathbf{k}\right) \times\left(3 t \mathbf{i}-12 t^{2} \mathbf{j}-6 t^{-2} \mathbf{k}\right)\right)$$

Short Answer

Expert verified
Question: Determine the derivative of the cross product of the vector functions \(\mathbf{a}(t) = t^3 \mathbf{i} + 6\mathbf{j} - 2\sqrt{t} \mathbf{k}\) and \(\mathbf{b}(t) = 3t \mathbf{i} - 12t^2 \mathbf{j} - 6t^{-2} \mathbf{k}\). Answer: The derivative of the cross product of the given vector functions is \(\frac{d}{dt}(\mathbf{a} \times \mathbf{b}) = (72t^{-3} + 60t^{1.5})\mathbf{i} + (-6 + 9t^{0.5})\mathbf{j} + (-60t^4 - 18)\mathbf{k}\).

Step by step solution

01

Identify the vector functions

Given the problem, we have two vector functions: $$\mathbf{a}(t) = t^3 \mathbf{i} + 6\mathbf{j} - 2\sqrt{t} \mathbf{k}$$ $$\mathbf{b}(t) = 3t \mathbf{i} - 12t^2 \mathbf{j} - 6t^{-2} \mathbf{k}$$
02

Compute the cross product \(\mathbf{a}(t) \times \mathbf{b}(t)\)

Using the cross product formula, we compute the cross product between the two vector functions: $$(\mathbf{a} \times \mathbf{b}) = \left|\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ t^3 & 6 & -2\sqrt{t} \\ 3t & -12t^2 & -6t^{-2} \\ \end{matrix}\right|$$ We then expand the matrix to obtain the components of the cross product: $$(\mathbf{a} \times \mathbf{b}) = (6(-6t^{-2}) - (-2\sqrt{t})(-12t^2))\mathbf{i} - (t^3(-6t^{-2}) - (-2\sqrt{t})(3t))\mathbf{j} + (t^3(-12t^2) - 6(3t))\mathbf{k}$$ Simplify: $$(\mathbf{a} \times \mathbf{b}) = (-36t^{-2} + 24t^{2.5})\mathbf{i} + (-6t + 6t\sqrt{t})\mathbf{j} + (-12t^5 - 18t)\mathbf{k}$$
03

Compute the derivative of the cross product

Now we need to find the derivative of the cross product with respect to \(t\). We compute the derivative for each component separately: $$\frac{d}{dt}(\mathbf{a} \times \mathbf{b}) = \frac{d}{dt}((-36t^{-2} + 24t^{2.5})\mathbf{i} + (-6t + 6t\sqrt{t})\mathbf{j} + (-12t^5 - 18t)\mathbf{k})$$ Derivative of each component: $$\frac{d}{dt}(-36t^{-2} + 24t^{2.5}) = 72t^{-3} + 60t^{1.5}$$ $$\frac{d}{dt}(-6t + 6t\sqrt{t}) = -6 + 9t^{0.5}$$ $$\frac{d}{dt}(-12t^5 - 18t) = -60t^4 - 18$$
04

Combine the derivatives

Now we can combine the derivatives of each component to find the total derivative of the cross product: $$\frac{d}{dt}(\mathbf{a} \times \mathbf{b}) = (72t^{-3} + 60t^{1.5})\mathbf{i} + (-6 + 9t^{0.5})\mathbf{j} + (-60t^4 - 18)\mathbf{k}$$ This is the derivative of the cross product with respect to \(t\).

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Most popular questions from this chapter

Evaluate the following definite integrals. $$\int_{1}^{4}\left(6 t^{2} \mathbf{i}+8 t^{3} \mathbf{j}+9 t^{2} \mathbf{k}\right) d t$$

Prove the following vector properties using components. Then make a sketch to illustrate the property geometrically. Suppose \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in the \(x y\) -plane and a and \(c\) are scalars. $$(a+c) \mathbf{v}=a \mathbf{v}+c \mathbf{v}$$

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