How Does the Height of a Parachute Change with Horizontal Displacement?

JakePearson
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the height h(x) in meters above the ground of a parachute varies with her horizontal distplacement x in meters from a landing target on the ground as h(x) = 50sin^-1 (0.1x). what is the rate of change of h with respect to x at x = 6m?

i was wondering of someone could help me with this question :)
 
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What is the derivative of h(x) = 50sin^-1 (0.1x)? That's all this is asking. Do you know the derivative of arcsine?
 
is it
1 / sqrt(1-x^2)

:)
 
JakePearson said:
is it
1 / sqrt(1-x^2)

:)


yes so now you know

\frac{d}{dx} (sin^{-1}x)= \frac{1}{\sqrt{1-x^2}}

what is d/dx( sin-1(0.1x)) ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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