# Check if answers are right on my review? (vector eqns, curvature )

1. Mar 31, 2013

### oreosama

Hi these are questions from my test review that i am unsure of, i posted question and my answer

if you can tell me if ive gotten right answer that would be much appreciated!

Let C be the curve with the equations $x = 2 - t^3, y = 2t - 6, z = \ln(t)$

Find the point where C intersects the xz-plane
Find parametric equations of the tangent line at (1,-4,0)

ans:
$(-25,0,ln(3))$

$x = 1 - 3t, y = -4 + 2t, z = t$

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find an equation of the osculating plane of the curve $x = \sin{5t}, y = \sqrt{5}t, z = \cos{5t}$ at the point $(0,\pi \sqrt{5}, -1)$

ans:

$-\frac{\sqrt{6}}{6} \cos{5t} x + \frac{\sqrt{30}}{6} ( y - \frac{\pi \sqrt{5}}{5}) + \frac{\sqrt{6}}{6} \sin{5t} (z + 1) = 0$

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Find the curvature of the curve $y = 2 \sqrt{x}$ at the point $(3, 2\sqrt{3})$

ans:

$\kappa(3) = \frac{1}{16}$

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An athlete throws a shot at an angle of 45 degrees to the horizontal at an initial speed of 36 ft/sec. It leaves his hand 4 feet above the ground.

Where is the shot 2 seconds later?
Where does the shot land?

ans:

$(36\sqrt{2} ft, 36\sqrt{2} - 15.6 ft)$

x = 132ft

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Find the tangential and normal components of the acceleration vector of a particle with position function $r(t) = \cos{t} i + \sin{t} j + \sqrt{15}t k$

ans:

at = 0
an = 1

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True/false...

The curve $r(t) = <0,t^2, 4t>$ is a parabola

T

The curve with the vector equation $r(t) = t^3 i + 2t^3j + 3t^3 k$ is a line

T

The binormal vector is $B(t) = N(t) x T(t)$

If curvature $\kappa(t) = 0$ for all t, the curve is a straight line

T

The curve $r(t) = <2t, 3 - t, 0 >$ is a line that passes through the origin.

F

If $|r(t)| = 1$ for all t, the r'(t) is orthogonal to r(t) for all t

T

if u(t) and v(t) are differentiable vector functions then $\frac{\delta}{\delta t}[u(t) x v(t)] = u'(t) x v'(t)$

F

I'm pretty sure the majority of these are right. the physics one and oscillating plane are the ones im kind of unsure of! thanks for any help i have test soon!

Last edited: Mar 31, 2013
2. Apr 1, 2013

### HallsofIvy

This can't be correct. There cannot be a "t" in the equation.