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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

The domain of the vector function

The vector function is given by:
r(t)=t2i+ln(2t)j+3k \mathbf{r}(t) = t^2 \mathbf{i} + \ln(2 - t) \mathbf{j} + 3 \mathbf{k}

The domain is specified to be the interval:

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Solution

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 logarithm is only defined for positive numbers).

Solving the inequality 2-t > 0 gives t < 2.

Therefore, the domain of the vector function is given by the interval (-∞, 2).

This problem has been solved

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.

Does the vector function F––(t)=ti–+2tcostj–+2tsintk––𝐹_(𝑡)=𝑡𝑖_+2𝑡𝑐𝑜𝑠𝑡𝑗_+2𝑡𝑠𝑖𝑛𝑡𝑘_ lie on a paraboloid?

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+.

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