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Chapter 11: Q. 63 (page 873)

Let k be a scalar and $r (t)$ be a differentiable vector function. Prove that $\frac{d}{d t} (k r (t)) = k r' (t)$ . (This is Theorem 11.11 (a).)

Short Answer

Expert verified

Ans: $\frac{d}{d t} (k 鉄� x (t), y (t), z (t) 鉄�) = k r^{'} (t)$

Step by step solution

01

Step 1. Given information:

k is scalar and

$r (t)$ be a differentiable vector function

02

Step 2. Proving:

Consider,

$\frac{d}{d t} (k r (t)) = \frac{d}{d t} (k 鉄� x (t), y (t), z (t) 鉄�) = \frac{d}{d t} 鉄� k x (t), k y (t), k z (t) 鉄� = <\frac{d}{d t} (k x (t)), \frac{d}{d t} (k y (t)), \frac{d}{d t} (k z (t))> = <k x^{'} (t), k y^{'} (t), k z^{'} (t)> = k <x^{'} (t), y^{'} (t), z^{'} (t)> = k r^{'} (t) T h u s, \frac{d}{d t} (k r (t)) = k r^{'} (t)$

Note: Let $u$ be a function of $x$ whose derivates exist.

Then $\frac{d}{d x} (k u) = k \frac{d u}{d x}$

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Most popular questions from this chapter

Given a vector-valued function r(t) with domain $鈩�,$ what is the relationship between the graph of r(t) and the graph of r(kt), where k > 1 is a scalar?

Let $r_{1} (t)$ and $r_{2} (t)$ both be differentiable three-component vector functions. Prove that

\frac{d}{d t} (r_{1} (t) 脳 r_{2} (t)) = r_{1}^{'} (t) 脳 r_{2} (t) + r_{1} (t) 脳 r_{2}^{'} (t)

(This is Theorem 11.11 (d).)

For each of the vector-valued functions, find the unit tangent vector.

$r (t) = (3 \sin t, 5 \cos t, 4 \sin t)$

Given a twice-differentiable vector-valued function $r (t)$ , what is the definition of the binormal vector $B (t)$ ?

Evaluate and simplify the indicated quantities in Exercises 35鈥�41.
$(1, t, t^{2}) - (t, t^{2}, t^{3})$

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