(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 ?

(a)

(b)

r'(5)= (10,75)

(c)

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)    [  and   ]

To find the , we put

        =(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

Differentiating with respect to t

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

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

6.5

step-by-step explanation:

you said it was in the question

~courtney