Question

Find the derivative r '(t) of the vector function r(t). t sin 6t , t2, t cos 7t Part 1 of 4 The derivative of a vector function is obtained by differentiating each of its components. Thus, if r(t) = f(t), g(t), h(t) , where f, g, and h are differentiable functions, then r'(t) = f '(t), g'(t), h'(t) . For r(t) = t sin 6t, t2, t cos 7t , we have f(t) = t sin 6t, which is a product. Using the Product Rule and the Chain Rule, we have

Answer 1

Answer:

$r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))$

Step-by-step explanation:

We need to find the derivative $r'(t)$ of the vector function :

$r(t)=(tsin(6t),t^{2},tcos(7t))$

In order to find $r'(t)$, we are going to differentiate each of its components ⇒

We can write the following ⇒

$r(t)=(f(t),g(t),h(t))=(tsin(6t),t^{2},tcos(7t))$ ⇒

$f(t)=tsin(6t)\\g(t)=t^{2}\\h(t)=tcos(7t)$

Let's differentiate each function to obtain $r'(t)$ :

$f(t)=tsin(6t)$ ⇒ $f'(t)=1.sin(6t)+t.cos(6t).6=sin(6t)+6tcos(6t)$ ⇒

$f'(t)=sin(6t)+6tcos(6t)$

Now with $g(t)$ :

$g(t)=t^{2}$ ⇒

$g'(t)=2t$

With $h(t)$ :

$h(t)=tcos(7t)$ ⇒ $h'(t)=1.cos(7t)+t[-sin(7t)].7$ ⇒

$h'(t)=cos(7t)-7tsin(7t)$

Finally we need to complete $r'(t)=(f'(t),g'(t),h'(t))$ with its components :

$r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))$