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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

Respuesta :

LucianoBordoli

Answer:

[tex]r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))[/tex]

Step-by-step explanation:

We need to find the derivative [tex]r'(t)[/tex] of the vector function :

[tex]r(t)=(tsin(6t),t^{2},tcos(7t))[/tex]

In order to find [tex]r'(t)[/tex], we are going to differentiate each of its components ⇒

We can write the following ⇒

[tex]r(t)=(f(t),g(t),h(t))=(tsin(6t),t^{2},tcos(7t))[/tex] ⇒

[tex]f(t)=tsin(6t)\\g(t)=t^{2}\\h(t)=tcos(7t)[/tex]

Let's differentiate each function to obtain [tex]r'(t)[/tex] :

[tex]f(t)=tsin(6t)[/tex] ⇒ [tex]f'(t)=1.sin(6t)+t.cos(6t).6=sin(6t)+6tcos(6t)[/tex] ⇒

[tex]f'(t)=sin(6t)+6tcos(6t)[/tex]

Now with [tex]g(t)[/tex] :

[tex]g(t)=t^{2}[/tex] ⇒

[tex]g'(t)=2t[/tex]

With [tex]h(t)[/tex] :

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

[tex]h'(t)=cos(7t)-7tsin(7t)[/tex]

Finally we need to complete [tex]r'(t)=(f'(t),g'(t),h'(t))[/tex] with its components :

[tex]r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))[/tex]

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