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balandron [24]
3 years ago
15

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

Mathematics
1 answer:
algol133 years ago
5 0

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}

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