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2 years ago

1. What is the difference between the altitude rule and the leg rule?

Mathematics

1 answer:

Anna71 [15]2 years ago

6 0

Answer:

The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle

The Leg Rule (or Leg geometric mean theorem) relates the length of each leg of a right triangle with the segments projected by them on the hypotenuse.

Step-by-step explanation:

The altitude to the hypotenuse of a right triangle is the mean proportional between the segments into which it divides the hypotenuse. Notice the triangle used with this rule. The leg of a right triangle is the mean proportional between the hypotenuse and the projection of the leg on the hypotenuse.

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2 years ago

The position function of a particle moving along a coordinate line is given, wheresis in feet andtis in seconds.

Sati [7]

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a) s = (4-t)/(t^2+4)^2, a(t) = (2t^3-24t)/(t^2+4)^3

b) s = 0.2ft, v = 0.12 ft/s, a = -0.176 ft/s^2

c) t = 2s

d) slowing down for t < 2, speeding up for t > 2

e) 0.327 ft

Step-by-step explanation:

The position function of a particle is given by:

$s(t)=\frac{t}{t^2+4},\ \ \ t\geq 0$ (1)

a) The velocity function is the derivative, in time, of the position function:

$v(t)=\frac{ds}{dt}=\frac{(1)(t^2+4)-t(2t)}{(t^2+4)^2}=\frac{4-t^2}{(t^2+4)^2}$ (2)

The acceleration is the derivative of the velocity:

$a(t)=\frac{dv}{dt}=\frac{(-2t)(t^2+4t)^2-(4-t^2)2(t^2+4)(2t)}{(t^2+4)^4}\\\\a(t)=\frac{(-2t)(t^2+4)-4t(4-t^2)}{(t^2+4)^3}=\frac{2t^3-24t}{(t^2+4)^3}$ (3)

b) For t = 1 you have:

$s(1)=\frac{1}{1+4}=0.2\ ft\\\\v(1)=\frac{4-1}{(1+4)^2}=0.12\frac{ft}{s}\\\\a(1)=\frac{2-24}{(1+4)^3}=-0.176\frac{ft}{s^2}$

c) The particle stops for v(t)=0. Then you equal equation (2) to zero ans solve the equation for t:

$v(t)=\frac{4-t^2}{(t^2+4)^2}=0\\\\4-t^2=0\\\\t=2$

For t = 2s the particle stops.

d) The second derivative evaluated in t=2 give us the concavity of the position function.

$\frac{d^2s}{dt^2}=a(2)=\frac{2(2)^3-24(2)}{(2^2+4)^3}=-0.062$

Then, the concavity of the position function is negative. For t=2 there is a maximum. Before t=2 the particle is slowing down and after t=2 the particle is speeding up.

e) Due to particle goes and come back. You first calculate s for t=2, then calculate for t=5.

$s(2)=\frac{2}{2^2+4}=0.25\ ft$

$s(5)=\frac{5}{5^2+4}=0.172\ ft$

The particle travels 0.25 in the first 2 seconds. In the following three second the particle comes back to the 0.172\ ft. Then, in the second trajectory the particle travels:

0.25 - 0.127 = 0.077 ft

The total distance is the sum of the distance of the two trajectories:

s_total = 0.25 ft + 0.077 ft = 0.327 ft

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