Answered

Find the derivative of the vector function r(t)=ta×(b+tc)r(t)=ta×(b+tc), where a=⟨5,3,−4⟩a=⟨5,3,−4⟩, b=⟨−4,1,1⟩b=⟨−4,1,1⟩, and c=⟨−3,2,−4⟩c=⟨−3,2,−4⟩.

Answer :

LammettHash

By the product rule,

[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\dfrac{\mathrm d(t\vec a)}{\mathrm dt}\times(\vec b+t\vec c)+t\vec a\times\dfrac{\mathrm d(\vec b+t\vec c)}{\mathrm dt}[/tex]

We have

[tex]t\vec a=\langle5t,3t,-4t\rangle\implies\dfrac{\mathrm d(t\vec a)}{\mathrm dt}=\langle5,3,-4\rangle[/tex]

[tex]\vec b+t\vec c=\langle-4-3t,1+2t,1-4t\rangle\implies\dfrac{\mathrm d(\vec b+t\vec c)}{\mathrm dt}=\langle-3,2,-4\rangle[/tex]

So

[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle5,3,-4\rangle\times\langle-4-3t,1+2t,1-4t\rangle+\langle5t,3t,-4t\rangle\times\langle-3,2,-4\rangle[/tex]

[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle7-4t,32t-21,19t-7\rangle+\langle-4t,32t,19t\rangle[/tex]

[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle7-8t,64t-21,38t-7\rangle[/tex]

Other Questions