Increasing Function: Finding k When x=pi/4

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Homework Statement


When x=pi/4, the rate at which sinx^2 is increasing is k times the rate at which x is increasing. What is the value of k?
(answer is 1)


Homework Equations





The Attempt at a Solution


I plugged in pi/4 into sinx^2 and i got .5.
i also took the derivative of sinx^2 and plugged in pi/4 to get the slope, which is 1.
So it seems to me that I found the rate at which sinx^2 is increasing (1), but i don't know how to find the rate at which x is increasing.
do i just take an average slope?
 
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Hint: (dy/dt) / (dx/dt) = dy/dx.
 
i think dy/dt would be 1.
how do i find dx/dt?
 
dx/dt is not given and is not needed. You only know that the ratio dy/dt and dx/dt is k. But the ratio of dy/dt and dx/dt is dy/dx, so what does that tell you?
 
that it is just simply the slope of sinx^2 when x=pi/4
which is 1
?
 
Yes, that's right. :smile:
 
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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