Question

(1 point) For the given position vectors r(t)r(t), compute the (tangent) velocity vector r′(t)r′(t) for the given value of tt . A) Let r(t)=(cos4t,sin4t)Let r(t)=(cos⁡4t,sin⁡4t). Then r′(π4)r′(π4)= ( , )? B) Let r(t)=(t2,t3)Let r(t)=(t2,t3). Then r′(5)r′(5)= ( , )? C) Let r(t)=e4ti+e−5tj+tkLet r(t)=e4ti+e−5tj+tk. Then r′(−5)r′(−5)= i+i+ j+j+ kk ?

Answer 1

Answer:

(a)

$r'(\frac \pi 4) =(0.-4)$

(b)

r'(5)= (10,75)

(c)

$r'(-5) =4 e^{-20}\hat i-5e^{25}\hat j+\hat k$

Step-by-step explanation:

(a)

Give that,the position vector is

r(t) = (cos 4t, sin 4t)

Differentiating with respect to t

r'(t) = (-4sin 4t, 4 cos 4t)    [$\frac{d}{dt} cos mt = -m \ sin \ mt$  and   $\frac{d}{dt} sin mt = m \ cos \ mt$]

To find the $r'(\frac\pi 4)$, we put $t=\frac \pi4$

$r'(\frac\pi 4) = (-4sin (4.\frac \pi 4), 4 cos (4.\frac \pi 4))$

        =(0, -4)

(b)

Give that,the position vector is

r(t) = (t²,t³)

Differentiating with respect to t

r'(t) = (2t, 3t²)

To find r'(5) ,  we put t=5

r'(5) = (2.5,3.5²)

      = (10,75)

(c)

Given position vector is

$r(t) = e^{4t}\hat i+e^{-5t}\hat j+t\hat k$

Differentiating with respect to t

$r'(t) =4 e^{4t}\hat i+(-5)e^{-5t}\hat j+\hat k$

$\Rightarrow r'(t) =4 e^{4t}\hat i-5e^{-5t}\hat j+\hat k$

To find r'(-5) ,  we put t= - 5 in the above equation

$r'(-5) =4 e^{4.(-5)}\hat i-5e^{-5.(-5)}\hat j+\hat k$

$\Rightarrow r'(-5) =4 e^{-20}\hat i-5e^{25}\hat j+\hat k$

Answer 2

Answthe answer is er

Step-by-step explanation: