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Ira Lisetskai [31]
3 years ago
14

Susan throws a softball upward into the air at a speed of 32 feet per second from a 88 dash foot88-foot platform. the distance u

pward that the ball travels is given by the function d left parenthesis t right parenthesis equals negative 16 t squared plus 32 t plus 88d(t)=−16t2+32t+88. what is the maximum height of the​ softball? how many seconds does it take to reach the ground after first being thrown​ upward? ​ (round your answer to the nearest​ tenth.)
Mathematics
1 answer:
Rudiy273 years ago
3 0
D(t)=-16t²+32t+88
a] Maximum height
At maximum height d'(t)=0
from d(t);
d'(t)=-32t+32=0
thus
t=1
hence the maximum height will be:
d(t)=-16t²+32t+88
d(1)=-16(1)^2+32(1)+88
d(1)=104 ft

2] <span> how many seconds does it take to reach the ground after first being thrown​ upward? ​
Here we solve for t intercept
</span>when d(t)=0
<span>thus</span>
0=-16t²+32t+88
<span>solving for t, we get:
t=1-</span>√13/2
or
<span>t=1+</span>√13/2
<span>
t=-1.55 or t=3.55 sec
Thus the answer is t=3.55 sec


</span>
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Let x = \arcsin(y), so that

\sin(x) = y

\tan(x)=\dfrac y{\sqrt{1-y^2}}

dx = \dfrac{dy}{\sqrt{1-y^2}}

Then the integral transforms to

\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}

\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy

Integrate by parts, taking

u = \ln(y) \implies du = \dfrac{dy}y

dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|

For 0 < y < 1, we have |1 - y²| = 1 - y², so

\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy

Recall the Taylor series for ln(1 + y),

\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n

Replacing y with -y² gives the Taylor series

\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}

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Interchanging the integral and sum (see Fubini's theorem) gives

\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy

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\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6

So, the value of our integral is

\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}

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