4 Calculus Problems

<ol>
<li><p>Find f'(x) and determine those values for which f'(x)=0
f(x)= ((3x+1)/(x^2))^2</p></li>
<li><p>Find all values of x for which the tangent to f(x)=1-sin2x is horizontal</p></li>
<li><p>A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15cm/min. At what rate must air be removed when the radius is 9 cm?</p></li>
<li><p>A man 6 ft tall is walking at the rate of 3ft/sec toward a street light 18ft tall.
(a) at what rate is the length of the shadow changing?
(b) how fast is the tip of his shadow moving?</p></li>
</ol>

<p>bump... HELP!</p>

<p>For # 1 use the quotient rule and then plug 0 into your answer.
For # 2 find the derivative and set that equal to zero.</p>

<ol>
<li><p>use chain rule/quotient rule. 2((3x+1)/(x^2)) * ((-3x^2-2x))/x^4. I simplified it and got [2(3x+1)(-3x-2)]/x^5. Then set it to 0. obviously whenever the numerator will = 0. So the answers are x=-1/3 and x=-2/3</p></li>
<li><p>f'(x)= -2cos2x. Set that = to 0 because we need the slope to equal 0. cos2x = 0. That makes x=pi/4, x=3pi/4</p></li>
</ol>

<ol>
<li>A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15cm/min. At what rate must air be removed when the radius is 9 cm?</li>
</ol>

<p>find dv/dt when r=9... given: dr/dt=15... im using p for pi
so v=4/3pr<em>3 so dv/dt=4pr</em>2dr/dt
9=4p(81)(15) and solve down</p>

<ol>
<li>A man 6 ft tall is walking at the rate of 3ft/sec toward a street light 18ft tall.
(a) at what rate is the length of the shadow changing?
(b) how fast is the tip of his shadow moving?</li>
</ol>

<p>this one use proportion and pythagareon theorem</p>

<ol>
<li>A man 6 ft tall is walking at the rate of 3ft/sec toward a street light 18ft tall.
(a) at what rate is the length of the shadow changing?
(b) how fast is the tip of his shadow moving?</li>
</ol>

<p>For this one, you just have to use similar triangles. </p>

<p>~~ |. +
~~ |.......+
~~ |...........+
~18| .............. +
~~ |..........6ft.|.....+<br>
~~ |________ |______+
~~ .......x...........s </p>

<p>theres my ghetto drawing of the situation.... </p>

<p>so first set up the proportion since you know there are 2 similar triangles:</p>

<p>(18)/(x+s) = (6)/(s) ...cross multiply and you get: 18s = 6x + 6s</p>

<p>simplify and you get: 2s = x</p>

<p>now just differential this in terms of time: d(2s)/dt = d(x)/dt
and you get: 2* ds/dt = dx/dt
or: ds/dt (which is what we're trying to find) = dx/dt divided by 2</p>

<p>and we know that dx/dt = -3 so ds/dt = -3/2 ft/sec ...thats how fast the length is moving ...so </p>

<p>a.) its shrinking at 1.5ft/sec.</p>

<p>and for b.) its just how fast its shrinking (1.5ft/sec) added to how fast the guy is actually moving ...so 1.5 + 3 = 4.5 ft/sec</p>

<p>thanks so much<32333</p>

<p>oops #2 is f(x)= x-sin2x ... not 1</p>

<p>also, ughstinkysocks, for #3 theres nothing to solve down. i know the answer is supposed to be 4860 i just dont know how to get it.</p>

<h1>2 just do the derivative. f'(x) = 1-2cos2x. Then set it to 0 for the slope.</h1>

<p>1-2cos2x = 0</p>

<p>cos2x = 1/2</p>

<p>x=pi/6</p>