Question

For the given position vectors r(t) compute the unit tangent vector T(t) for the given value of t .

Answer 1

Answer:

a) $T(t) = \frac{}{5}=$

$T(4) =$

b) $T(t) = \frac{}{8\sqrt{37}}$

$T(4) =$

c) $T(t) = \frac{}{2425825977}$

$T(4) = \frac{1}{2425825977}$

Step-by-step explanation:

The tangent vector is defined as:

$T(t) = \frac{r'(t)}{|r'(t)|}$

Part a

For this case we have the following function given:

$r(t) =$

The derivate is given by:

$r'(t) =$

The magnitude for the derivate is given by:

$|r'(t)| = \sqrt{25 sin^2(5t) +25 cos^2 (5t)}= 5\sqrt{cos^2 (5t) + sin^2 (5t)} =5$

And then the tangent vector for this case would be:

$T(t) = \frac{}{5}=$

And for the case when t=4 we got:

$T(4) =$

Part b

For this case we have the following function given:

$r(t) =$

The derivate is given by:

$r'(t) =$

The magnitude for the derivate is given by:

$|r'(t)| = \sqrt{4t^2 +9t^4}= t\sqrt{4 + 9t^2}$

$|r'(4)| = \sqrt{4(4)^2 +9(4)^4}= 4\sqrt{4 + 9(4)^2} = 4\sqrt{148}= 8\sqrt{37}$

And then the tangent vector for this case would be:

$T(t) = \frac{}{8\sqrt{37}}$

And for the case when t=4 we got:

$T(4) =$

Part c

For this case we have the following function given:

$r(t) =$

The derivate is given by:

$r'(t) =$

The magnitude for the derivate is given by:

$|r'(t)| = \sqrt{25e^{10t} +16e^{-8t} +1}$

$|r'(t)| = \sqrt{25e^{10*4} +16e^{-8*4} +1} =2425825977$

And then the tangent vector for this case would be:

$T(t) = \frac{}{2425825977}$

And for the case when t=4 we got:

$T(4) = \frac{1}{2425825977}$