Exam 13: Vector Functions

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Find parametric equations for the tangent line to the curve with parametric equations x=3tx = 3 t , y=5t2,z=4t3y = 5 t ^ { 2 } , z = 4 t ^ { 3 } at the point with t=1t = 1 .

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B

A projectile is fired with an initial speed of 700 m/s700 \mathrm {~m} / \mathrm { s } and angle of elevation 6060 ^ { \circ } . Find the range of the projectile.

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d43.3 kmd \approx 43.3 \mathrm {~km}

Find a vector function that represents the curve of intersection of the two surfaces: The circular cylinder x2+y2=4x ^ { 2 } + y ^ { 2 } = 4 and the parabolic cylinder z=xyz = x y .

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E

Sketch the curve of the vector function r(t)=t2,t3,t,t0\mathbf { r } ( t ) = \left\langle t ^ { 2 } , t ^ { 3 } , t \right\rangle , t \geq 0 , and indicate the orientation of the curve.

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The curves r1(t)=t,t3,t9}\mathbf { r } _ { 1 } ( t ) = \left\langle t , t ^ { 3 } , t ^ { 9 } \right\} and r2(t)=sint,sin5t,t\mathbf { r } _ { 2 } ( t ) = \langle \sin t , \sin 5 t , t \rangle intersects at the origin. Find their angle of intersection correct to the nearest degree.

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 Find rt(t) and rtt(t) for r(t)=10ti+7t2j+5t3k\text { Find } \mathbf { r } ^ { t } ( t ) \text { and } \mathbf { r } ^ {tt } ( t ) \text { for } \mathbf { r } ( t ) = 10 t \mathbf { i } + 7 t ^ { 2 } \mathbf { j } + 5 t ^ { 3 } \mathbf { k }

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Find the scalar tangential and normal components of acceleration of a particle with position vector r(t)=4sinti+4costj+3tk\mathbf { r } ( t ) = 4 \sin t \mathbf { i } + 4 \cos t \mathbf { j } + 3 t \mathbf { k }

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A particle moves with position function r(t)=(21t7t36)i+21t2j\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 6 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j } Find the tangential component of the acceleration vector. Select the correct answer.

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 If r(t)=7i+tcosπtj+4sinπtk, evaluate 01r(t)dt\text { If } \mathbf { r } ( t ) = 7 \mathbf { i } + t \cos \pi t \mathbf { j } + 4 \sin \pi t \mathbf { k } \text {, evaluate } \int _ { 0 } ^ { 1 } r ( t ) d t

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Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid x2+6y2+6z2=36x ^ { 2 } + 6 y ^ { 2 } + 6 z ^ { 2 } = 36 and the parabolic cylinder y=x2y = x ^ { 2 } .

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Find the scalar tangential and normal components of acceleration of a particle with position vector r(t)=et{cos4t,sin4t,0}\mathbf { r } ( t ) = e ^ { t } \{ \cos 4 t , \sin 4 t , 0 \}

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 Find the unit tangent vector T(t) for r(t)=4ti+2tj+4tk at t=2\text { Find the unit tangent vector } \mathrm { T } ( t ) \text { for } \mathbf { r } ( t ) = 4 t \mathbf { i } + 2 t \mathbf { j } + 4 t \mathbf { k } \text { at } t = 2 \text {. }

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Given r(t)=7t2i+4t3j\mathbf { r } ( t ) = 7 t ^ { 2 } \mathbf { i } + 4 t ^ { 3 } \mathbf { j } . a. Find r(1)\mathbf { r } ( - 1 ) and r(1)\mathbf { r } ^ { \prime } ( - 1 ) . b. Sketch the curve defined by r\mathbf { r } and the vectors r(1)\mathbf { r } ( - 1 ) and rt(1)\mathbf { r } ^ { t } ( - 1 ) on the same set of axes.

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Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid x2+7y2+7z2=49x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49 and the parabolic cylinder y=x2y = x ^ { 2 } .

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Find the length of the curve r(t)=2ti+t2j+lntk,1te3\mathbf { r } ( t ) = 2 t \mathbf { i } + t ^ { 2 } \mathbf { j } + \ln t \mathbf { k } , 1 \leq t \leq e ^ { 3 } .

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If r(t)=t,t9,t11}\mathbf { r } ( t ) = \left\langle t , t ^ { 9 } , t ^ { 11 } \right\} , find rtt(t)\mathbf { r } ^ { tt } ( t )

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 Find r(t) if rt(t)=t10i+5t4jt7k and r(0)=j\text { Find } r ( t ) \text { if } \mathbf { r } ^ { t } ( t ) = t ^ { 10 } \mathbf { i } + 5 t ^ { 4 } \mathbf { j } - t ^ { 7 } \mathbf { k } \text { and } \mathrm { r } ( 0 ) = \mathrm { j }

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The torsion of a curve defined by r(t)\mathbf { r } ( t ) is given by τ=(rt×rtt)rtttrt×rtt2\tau = \frac { \left( \mathbf { r } ^ { t } \times \mathbf { r } ^ { tt } \right) \cdot \mathbf { r } ^ {ttt } } { \left| \mathbf { r } ^ { t } \times \mathbf { r } ^ { t t } \right| ^ { 2 } } Find the torsion of the curve defined by r(t)=cos5ti+sin5tj+4tkr ( t ) = \cos 5 t \mathbf { i } + \sin 5 t \mathbf { j } + 4 t \mathbf { k } .

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Use Simpson's Rule with n=4\mathrm { n } = 4 to estimate the length of the arc of the curve with equations x=t,y=4t,z=t2+1x = \sqrt { t } , y = \frac { 4 } { t } , z = t ^ { 2 } + 1 , from (1,4,2)( 1,4,2 ) to (2,1,17)( 2,1,17 ) . Round your answer to four decimal places.

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 A particle moves with position function r(t)=(21t7t35)i+21t2j\text { A particle moves with position function } \mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j } \text {. }  Find the tangential component of the acceleration vector. \text { Find the tangential component of the acceleration vector. }

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