Find a tangent vector of unit length at the point with the given value of the parameter t. r(t) = (7 + t²)i + t²j, t = 1

Solution:

It is given that

r(t) = (7 + t²)i + t²j, t = 1

By differentiating it with respect to t

r’ (t) = 2t i + 2t j

Here the magnitude is

\(\sqrt{(2t)^{2}+(2t)^{2}}=\sqrt{8t^{2}}=2\sqrt{2}t\)

Using the value of t = 1

r’ (1) = 2 (1) i + 2 (1) j

r’ (1) = 2i + 2j

Here the magnitude is

\(\sqrt{(2)^{2}+(2)^{2}}=\sqrt{4^{2}}=2\sqrt{2}\)

So the tangent vector is

Tangent vector of unit length = First derivative of given point/ Magnitude of this derivative

= (2i + 2j)/ 2√2

= √2/2 i + √2/2 j

Therefore, the tangent vector of unit length is √2/2 i + √2/2 j.

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Find a tangent vector of unit length at the point with the given value of the parameter t. r(t) = (7 + t²)i + t²j, t = 1

Summary:

The tangent vector of unit length at the point with the given value of the parameter t r(t) = (7 + t²)i + t²j, t = 1 is √2/2 i + √2/2 j.