3.2 Calculus of vector-valued functions (Page 2/11)

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Differentiation of vector-valued functions

Let f, g, and h be differentiable functions of t.

If $r (t) = f (t) i + g (t) j,$ then $r^{'} (t) = f^{'} (t) i + g^{'} (t) j .$
If $r (t) = f (t) i + g (t) j + h (t) k,$ then $r^{'} (t) = f^{'} (t) i + g^{'} (t) j + h^{'} (t) k .$

Calculating the derivative of vector-valued functions

Use [link] to calculate the derivative of each of the following functions.

$r (t) = (6 t + 8) i + (4 t^{2} + 2 t - 3) j$
$r (t) = 3 cos t i + 4 sin t j$
$r (t) = e^{t} sin t i + e^{t} cos t j - e^{2 t} k$

We use [link] and what we know about differentiating functions of one variable.

The first component of $r (t) = (6 t + 8) i + (4 t^{2} + 2 t - 3) j$ is $f (t) = 6 t + 8 .$ The second component is $g (t) = 4 t^{2} + 2 t - 3 .$ We have $f^{'} (t) = 6$ and $g^{'} (t) = 8 t + 2,$ so the theorem gives $r^{'} (t) = 6 i + (8 t + 2) j .$
The first component is $f (t) = 3 cos t$ and the second component is $g (t) = 4 sin t .$ We have $f^{'} (t) = −3 sin t$ and $g^{'} (t) = 4 cos t,$ so we obtain $r^{'} (t) = −3 sin t i + 4 cos t j .$
The first component of $r (t) = e^{t} sin t i + e^{t} cos t j - e^{2 t} k$ is $f (t) = e^{t} sin t,$ the second component is $g (t) = e^{t} cos t,$ and the third component is $h (t) = - e^{2 t} .$ We have $f^{'} (t) = e^{t} (sin t + cos t),$ $g^{'} (t) = e^{t} (cos t - sin t),$ and $h^{'} (t) = −2 e^{2 t},$ so the theorem gives $r^{'} (t) = e^{t} (sin t + cos t) i + e^{t} (cos t - sin t) j - 2 e^{2 t} k .$

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Calculate the derivative of the function

r (t) = (t ln t) i + (5 e^{t}) j + (cos t - sin t) k .

$r^{'} (t) = (1 + ln t) i + 5 e^{t} j - (sin t + cos t) k$

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We can extend to vector-valued functions the properties of the derivative that we presented in the Introduction to Derivatives . In particular, the constant multiple rule , the sum and difference rules , the product rule , and the chain rule all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions: (1) for a real-valued function multiplied by a vector-valued function, (2) for the dot product of two vector-valued functions, and (3) for the cross product of two vector-valued functions.

Properties of the derivative of vector-valued functions

Let r and u be differentiable vector-valued functions of t , let f be a differentiable real-valued function of t, and let c be a scalar.

\begin{matrix} i. & \frac{d}{d t} [c r (t)] & = & c r^{'} (t) & Scalar multiple \\ ii. & \frac{d}{d t} [r (t) \pm u (t)] & = & r^{'} (t) \pm u^{'} (t) & Sum and difference \\ iii. & \frac{d}{d t} [f (t) u (t)] & = & f^{'} (t) u (t) + f (t) u^{'} (t) & Scalar product \\ iv. & \frac{d}{d t} [r (t) \cdot u (t)] & = & r^{'} (t) \cdot u (t) + r (t) \cdot u^{'} (t) & Dot product \\ v. & \frac{d}{d t} [r (t) \times u (t)] & = & r^{'} (t) \times u (t) + r (t) \times u^{'} (t) & Cross product \\ vi. & \frac{d}{d t} [r (f (t))] & = & r^{'} (f (t)) \cdot f^{'} (t) & Chain rule \\ vii. & If r (t) \cdot r (t) & = & c, then r (t) \cdot r^{'} (t) = 0 . \end{matrix}

Proof

The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule from the Introduction to Derivatives . Let $u (t) = g (t) i + h (t) j .$ Then

\begin{matrix} \frac{d}{d t} [f (t) u (t)] & = \frac{d}{d t} [f (t) (g (t) i + h (t) j)] \\ = \frac{d}{d t} [f (t) g (t) i + f (t) h (t) j] \\ = \frac{d}{d t} [f (t) g (t)] i + \frac{d}{d t} [f (t) h (t)] j \\ = (f^{'} (t) g (t) + f (t) g^{'} (t)) i + (f^{'} (t) h (t) + f (t) h^{'} (t)) j \\ = f^{'} (t) u (t) + f (t) u^{'} (t) . \end{matrix}

To prove property iv. let $r (t) = f_{1} (t) i + g_{1} (t) j$ and $u (t) = f_{2} (t) i + g_{2} (t) j .$ Then

\begin{matrix} \frac{d}{d t} [r (t) \cdot u (t)] & = \frac{d}{d t} [f_{1} (t) f_{2} (t) + g_{1} (t) g_{2} (t)] \\ = f_{1}^{'} (t) f_{2} (t) + f_{1} (t) f_{2}^{'} (t) + g_{1}^{'} (t) g_{2} (t) + g_{1} (t) g_{2}^{'} (t) \\ = f_{1}^{'} (t) f_{2} (t) + g_{1}^{'} (t) g_{2} (t) + f_{1} (t) f_{2}^{'} (t) + g_{1} (t) g_{2}^{'} (t) \\ = (f_{1}^{'} i + g_{1}^{'} j) \cdot (f_{2} i + g_{2} j) + (f_{1} i + g_{1} j) \cdot (f_{2}^{'} i + g_{2}^{'} j) \\ = r^{'} (t) \cdot u (t) + r (t) \cdot u^{'} (t) . \end{matrix}

The proof of property v. is similar to that of property iv. Property vi. can be proved using the chain rule. Last, property vii. follows from property iv:

\begin{matrix} \frac{d}{d t} [r (t) \cdot r (t)] & = & \frac{d}{d t} [c] \\ r^{'} (t) \cdot r (t) + r (t) \cdot r^{'} (t) & = & 0 \\ 2 r (t) \cdot r^{'} (t) & = & 0 \\ r (t) \cdot r^{'} (t) & = & 0 . \end{matrix}

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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