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marin [14]
3 years ago
9

Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mathematics
1 answer:
Mumz [18]3 years ago
4 0

Answer:

Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

Step-by-step explanation:

The equation of the curve is

Y = sin^{-1}(7x)

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

Y'=\frac{7}{\sqrt{1-49x^2} }

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is  x = \sqrt{\frac{1}{7} }

Then slope would accordingly be

Y'=\frac{7}{\sqrt{1-49/49} }

= ∞

For, x = \sqrt{\frac{1}{7} }, Y = sin^{-1}(7/7)= \pi/2

Equation of tangent line, in the point slope form, would be Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

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Answer:

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Step-by-step explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform Y\sim Unif(A,B). And the density function would be given by:

f(x) =\frac{1}{B-A} , A

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}

And we can find the probability desired like this:

P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)

And from the cumulative distribution function of X ficen by F(X)\frac{X-A}{B-A} we got:

\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}

Part b

For this case we assume that X\sim Gamma (2,1)

On this case we assume that \alpha=2, \beta= 1

The density function for the Gamma distribution is given by:

P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}

And on this case we can find the probability using the complement rule like this:

P(X>1) = 1-P(X\leq 1)=0.736

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736

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Answer:

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