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3 years ago

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A jumping spider's movement is modeled by a parabola. The spider makes a single jump from the origin and reaches a maximum heigh

t of 10 mm halfway across a horizontal distance of 80 mm.
Part A: Write the equation of the parabola in standard form that models the spider's jump. Show your work. (4 points)

Part B: Identify the focus, directrix, and axis of symmetry of the parabola. (6 points)

1 answer:

Anna11 [10]3 years ago

4 0

The spider's movement is an illustration of a parabola.

The equation of the parabola is: $\mathbf{y = -\frac{1}{320}(x - 80)^2 + 20}$
The focus of the parabola is $\mathbf{Focus = (80 ,-60)}$
The directrix is: $\mathbf{y = 100}$ .
The axis of symmetry is: $\mathbf{x = 80}$

<u>(a) The equation</u>

The spider passes through the origin.

So, we have:

$\mathbf{(x,y) = (0,0)}$

The spider jumps to a maximum height of 20mm, midway 160mm.

So, the vertex is:

$\mathbf{(h,k) = (80,20)}$

The equation of a parabola is:

$\mathbf{y = a(x - h)^2 + k}$

So, we have:

$\mathbf{0 = a(0 - 80)^2 + 20}$

$\mathbf{0 = 6400a + 20}$

Subtract 20 from both sides

$\mathbf{6400a =- 20}$

Solve for a

$\mathbf{a =- \frac{1}{320}}$

Substitute $\mathbf{a =- \frac{1}{320}}$ and $\mathbf{(h,k) = (80,20)}$ in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{y = -\frac{1}{320}(x - 80)^2 + 20}$

<u>(b) The focus, directrix and the axis of symmetry</u>

The focus of a parabola is:

$\mathbf{Focus = (h,k+p)}$

Where:

$\mathbf{p = \frac{1}{4a}}$

So, we have:

$\mathbf{p = \frac{1}{4\times -1/320}}$

$\mathbf{p = -\frac{320}{4}}$

$\mathbf{p = -80}$

So, we have:

$\mathbf{Focus = (80 , 20-80)}$

$\mathbf{Focus = (80 ,-60)}$

The axis of symmetry is:

$\mathbf{x = h}$

So, we have:

$\mathbf{x = 80}$

The directrix is:

$\mathbf{y = k - p}\\$

So, we have:

$\mathbf{y = 20+80}$

$\mathbf{y = 100}$

Read more about parabolas at:

brainly.com/question/5430838

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# # #

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$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

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Finally replace $u=\ln t$ to get the solution we found earlier,

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