MATHEMATICS COLLEGE

(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 ?

Answers

Answer 1

Answer:

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) [ $(d)/(dt) cos mt = -m \ sin \ mt$ and $(d)/(dt) sin mt = m \ cos \ mt$ ]

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

Answer:

For the given position vectors r(t)r(t), compute the (tangent) velocity vector r′(t)r′(t) for the given value of tt are:

$A) r' (\pi /4) = (0, -4) \nB) r'(5) = (10, 75)\nC) r'(-5) = (4e^(-20), -5e^(25), 1)$

To compute the velocity vector, we need to find the derivative of the position vector with respect to time (t). This will give us the tangent velocity vector.

A) Let r(t) = (cos⁡4t, sin⁡4t).

To find r'(t), we take the derivative of each component with respect to t:

r'(t) = (d/dt (cos⁡4t), d/dt (sin⁡4t))

r'(t) = (-4sin⁡4t, 4cos⁡4t)

To find r'(π/4), we substitute t = π/4 into r'(t):

r'(π/4) = (-4sin⁡(4(π/4)), 4cos⁡(4(π/4)))

r'(π/4) = (-4sin⁡π, 4cos⁡π)

r'(π/4) = (0, -4)

B) $Let \ r(t) = (t^2, t^3).$

To find r'(t), we take the derivative of each component with respect to t:

$r'(t) = (d/dt (t^2), d/dt (t^3))\nr'(t) = (2t, 3t^2)$

To find r'(5), we substitute t = 5 into r'(t):

$r'(5) = (2(5), 3(5)^2)\nr'(5) = (10, 75)$

C) Let $r(t) = e^(4t)i + e^(-5t)j + tk.$

To find r'(t), we take the derivative of each component with respect to t:

$r'(t) = (d/dt (e^(4t)), d/dt (e^(-5t)), d/dt (t))]\n\nr'(t) = (4e^(4t)), -5e^(-5t), 1)$

To find r'(-5), we substitute t = -5 into r'(t):

$r'(-5) = (4e^(4{-5}), -5e^(-5(-5)), 1) \n\nr'(-5) = (4e^(-20), -5e^(25), 1)$

So, the answers are:

$A) r' (\pi /4) = (0, -4) \nB) r'(5) = (10, 75)\nC) r'(-5) = (4e^(-20), -5e^(25), 1)$

To know more about vectors:

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Answers

Answer:

Step-by-step explanation:

x + y = 5 => x = 5 - y

y - 4 = - x

=> y - 4 = - ( 5 - y)

=> y - 4 = - 5 + y

=>4 = 5 , not true

So, no solution => system is inconsistent

option b

Explanation just as above.

Line p is parallel to line q. Parallel lines p and q are crossed by lines a and b to form 2 triangles. At parallel line p, angle 4 is formed by line b and angle 5 is formed by line a. Angle 6 is the third angle. At parallel line q, angle 3 is formed by line 3 and angle 2 is formed by line b. Angle 1 is the third angle. Which set of statements about the angles is true? Angle 1 is congruent to angle 6, angle 5 is congruent to angle 4, angle 3 is congruent to angle 2 Angle 2 is congruent to angle 4, angle 3 is congruent to angle 6, angle 1 is congruent to angle 5 Angle 3 is congruent to angle 6, angle 1 is congruent to angle 2, angle 5 is congruent to angle 4 Angle 6 is congruent to angle 1, angle 5 is congruent to angle 3, angle 4 is congruent to angle 2

Answers

Answer:

Step-by-step explanation:

Answer:

add all angles

Step-by-step explanation:

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