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)) ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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