Question

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


A) Let r(t)=(cos(5t),sin(5t)).

Then T(?4)= (___,___)

B) Let r(t)=(t2,t3).

Then T(4)= (___,___)

C) Let r(t)=e5ti+e?4tj+tk.

Then T(?4)=__i+__j+__k

Answer 1

Answer:

a) $T(t) = \frac{<-5 sin(5t), 5cos(5t)>}{5}= <-sin(5t), cos(5t)>$

$T(4) = <-sin(20), cos(20)>$

b) $T(t) = \frac{<t^2, 3t^2>}{8\sqrt{37}}$

$T(4) = <\frac{2\sqrt{37}}{37},\frac{6\sqrt{37}}{37} >$

c) $T(t) = \frac{<5e^{5t}, -4e^{-4t}, 1>}{2425825977}$

$T(4) = \frac{1}{2425825977}<5e^{50}, -4e^{-16},1 >$

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:

$r(t) = <cos(5t), sin(5t)>$

The derivate is given by:

$r'(t) = <-5 sin(5t), 5cos(5t)>$

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:

$T(t) = \frac{<-5 sin(5t), 5cos(5t)>}{5}= <-sin(5t), cos(5t)>$

And for the case when t=4 we got:

$T(4) = <-sin(20), cos(20)>$

Part b

For this case we have the following function given:

$r(t) = <t^2, t^3>$

The derivate is given by:

$r'(t) = <2t, 3t^2>$

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:

$T(t) = \frac{<t^2, 3t^2>}{8\sqrt{37}}$

And for the case when t=4 we got:

$T(4) = <\frac{2\sqrt{37}}{37},\frac{6\sqrt{37}}{37} >$

Part c

For this case we have the following function given:

$r(t) = <e^{5t}, e^{-4t} ,t>$

The derivate is given by:

$r'(t) = <5e^{5t}, -4e^{-4t}, 1>$

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:

$T(t) = \frac{<5e^{5t}, -4e^{-4t}, 1>}{2425825977}$

And for the case when t=4 we got:

$T(4) = \frac{1}{2425825977}<5e^{50}, -4e^{-16},1 >$