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Svetradugi
3 months 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:
Svet_ta [12.7K]3 months 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 determined as follows:

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

Part a

The given function for this instance is:

r(t) =

The derivative is calculated as:

r'(t) =

The magnitude of the derivative is determined by:

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

Thus, the tangent vector is found to be:

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

Finally, for t=4 we find:

T(4) =

Part b

In this case, we have the stated function:

r(t) =

The derivative can be expressed as:

r'(t) =

The magnitude for the derivative is calculated as:

|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}

Thus, the tangent vector in this case is:

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

Finally, when t=4 we find:

T(4) =

Part c

Here, we have the corresponding function stated:

r(t) =

The derivative is stated as:

r'(t) =

The magnitude for the derived value is calculated as:

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

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

Therefore, we conclude the tangent vector here:

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

And for the case when t=4, we get:

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

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