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MissTica
3 years ago
5

A particle is moving along the parabola x^2=4(y+6). As the particle passes through the point (8,10), the rate of change of its y

-coordinate is 5 units per second. How fast, in units per second, is the x-coordinate changing at this instant?
Mathematics
1 answer:
Fofino [41]3 years ago
8 0
You'll need to use differentiation (specifically, implicit differentiation) here.

If x^2 = 4(y+6), differentiating both sides with respect to time t produces the following:

2x (dx/dt) = 4([dy/dt])   (note that (d/dt) 6 = 0)

We need to solve for (dx/dt).  Substitute 8 for x (y does not appear in this latest equation, so we do nothing with y=10).  Substitute the given 5 units/sec for dy/dt:

2(8)(dx/dt) = 4(5)(units/sec)

Solving for dx/dt, dx/dt = [20 units/sec]/16, or 5/4 units/sec, or 1.25 units/sec.
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3 years ago
(a) Use a linear approximation to estimate f(0.9) and f(1.1). f(0.9) ≈ f(1.1) ≈ (b) Are your estimates in part (a) too large or
olasank [31]

Answer:

(Missing part of the question is attached)

L(x)=2x+3

Estimates are too large.

Step-by-step explanation:

Suppose the only information we know about the function is:

f(1)=5

where the graph of its derivative is shown in the attachment

<h3>(a)</h3>

If the function f\\ is differentiable at point x=1 , the tangent line to the graph of f at 1 is given by the equation:

y=f(1) +f'(1)(x-1)

So we call the linear function:

L(x)=f(1) +f'(1)(x-1)

We know the f(1)=5 as it is given in the question, and f'(1)=2 from the graph attached. Substitute in the equation of L(x).

L(x)=5+2(x-1)\\L(x)=5+2x-2\\L(x)=2x+3\\

<h3>(b)</h3>

At x=1,  f'(x) is positive but it is decreasing. However. if we draw the tangent lines, we see that the tangent lines are becoming less steeper, so the tangent lines lie above the curve f. Thus, The estimates are too large.

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Answer:

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Step-by-step explanation:

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