The domain of the vector function r–(t)=t2i–+ln(2−t)j–+3k––𝑟_(𝑡)=𝑡2𝑖_+𝑙𝑛(2−𝑡)𝑗_+3𝑘_ is given by the interval
Question
Solution 1
The domain of the vector function r(t) = t^2i + ln(2-t)j + 3k is determined by the range of t for which the function is defined.
The first and third components, t^2i and 3k, are defined for all real numbers.
However, the second component, ln(2-t)j, is only defined for 2-t > 0 (since the natural Knowee AI is a powerful AI-powered study tool designed to help you to solve study problem.
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Similar Questions
The domain of the vector function r–(t)=t2i–+ln(2−t)j–+3k––𝑟_(𝑡)=𝑡2𝑖_+𝑙𝑛(2−𝑡)𝑗_+3𝑘_ is given by the interval
The domain of the vector function F––(t)=t−3−−−−√i–+ln(10−2t−−−−−−√)–j+1t−3√k––𝐹_(𝑡)=𝑡−3𝑖_+𝑙𝑛(10−2𝑡)_𝑗+1𝑡−3𝑘_ is given by the interval
Let F––(t)=t2i–+tj–+k––𝐹_(𝑡)=𝑡2𝑖_+𝑡𝑗_+𝑘_ and G––(t)=i–+tj–+t2k––𝐺_(𝑡)=𝑖_+𝑡𝑗_+𝑡2𝑘_ be two vector functions defined for each t∈R𝑡∈𝑅.
Let F––(t)=2ti–−5j–+t2k––𝐹_(𝑡)=2𝑡𝑖_−5𝑗_+𝑡2𝑘_ and G––(t)=(1−t)i–+1tk––𝐺_(𝑡)=(1−𝑡)𝑖_+1𝑡𝑘_ where t∈R+𝑡∈𝑅+.
A.𝑓(𝑡)=2𝑡−𝑡3f(t)=2 t −t 3 B.𝑑(𝑡)=(1.1)𝑡d(t)=(1.1) t C.𝑏(𝑡)=𝑡4−3𝑡+9b(t)=t 4 −3t+9D.ℎ(𝑡)=5𝑡+𝑡5h(t)=5 t +t 5 E.𝑐(𝑡)=𝑡2−5𝑡c(t)= t 2 −5t
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