How to do the following question?

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How to do the following question?

How to find: given $A = x \sqrt{100 - x^{2}}$ $A = x sqrt(100 - x^2)$ and $\frac{d x}{d t}$ $dx/dt$ = 0.3, the valuue of $\frac{d A}{d t}$ $(dA)/dt$ when x = 8 is?

A particle is moving along a curve with equation $y = x^{2}$ $y=x^2$ . If $\frac{d x}{d t} = 2$ $dx/dt = 2$ at $x = 3$ $x = 3$ , find $\frac{d y}{d t}$ $dy/dt$ at that instant.

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Answer 1 · 2018-07-06T08:04:39

Take into account that $\frac{d A}{d t} = \frac{d A}{d x} \cdot \frac{d x}{d t}$ $(dA)/dt=(dA)/dx·dx/dt$ (chain rule)

So, there is a functional relationship between x and t. In this case is linear

1) Then $\frac{d A}{d x} = \sqrt{100 - x^{2}} - \frac{x^{2}}{\sqrt{100 - x^{2}}}$ $(dA)/dx=sqrt(100-x^2)-(x^2)/(sqrt(100-x^2)$ at $x = 8$ $x=8$ his value is $\frac{d A}{d x} = 6 - \frac{64}{6} = - \frac{14}{3}$ $(dA)/dx=6-64/6=-14/3$

$\frac{d A}{d t} = \frac{d A}{d x} \cdot \frac{d x}{d t} = - \frac{14}{3} \cdot 0.3 = - \frac{7}{5}$ $(dA)/dt=(dA)/dx·dx/dt=-14/3·0.3=-7/5$ at $x = 8$ $x=8$

2) By the same reason above, $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t} = 2 x \cdot 2 = 12$ $(dy)/dt=(dy)/dx·(dx)/dt=2x·2=12$ at $x = 3$ $x=3$

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How to find: given A=x√100−x2A = x sqrt(100 - x^2) and dxdtdx/dt = 0.3, the valuue of dAdt(dA)/dt when x = 8 is? A particle is moving along a curve with equation y=x2y=x^2. If dxdt=2dx/dt = 2 at x=3x = 3, find dydtdy/dt at that instant.

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