1. ## Calc problems

Could anyone please help us out with these problems even just to show how to begin. It would be great and I would be very grateful. I have an exam coming up and I need to get these problems sorted.

Show that the volume of the solid obtained by rotating the part of circle of x^2 + y^2 = 1 that lies between x =0 and x = 1/2 about the x axis is (11pi)/24 cubic units.

Find the antiderivative between e and 1 of (4 ln(x) + 5) / x dx using substitution of w = 4ln(x) + 5

A searchlight rotates at a rate of 3 revolutions per min. Its beam of light hits a wall that is 10km away and produces a dot of light that moves horizontally along the wall. At what speed is the dot moving when the angle between the beam of light and the line through the searchlight perpendicular to the wall is pi/6

Find the slope of the curve x^2 + xy + 2y^3 = 4 at the point (-2,1)

2. Originally Posted by Mickey
Could anyone please help us out with these problems even just to show how to begin. It would be great and I would be very grateful. I have an exam coming up and I need to get these problems sorted.

Show that the volume of the solid obtained by rotating the part of circle of x^2 + y^2 = 1 that lies between x =0 and x = 1/2 about the x axis is (11pi)/24 cubic units.
first, $y = \pm \sqrt{1 - x^2}$ but because of the given boundary, you take the positive $y$.

use $V= \pi \int_0^{1/2} (\sqrt{1 - x^2})^2 \, dx$

3. Originally Posted by Mickey

Find the antiderivative between e and 1 of (4 ln(x) + 5) / x dx using substitution of w = 4ln(x) + 5

Originally Posted by Mickey
A searchlight rotates at a rate of 3 revolutions per min. Its beam of light hits a wall that is 10km away and produces a dot of light that moves horizontally along the wall. At what speed is the dot moving when the angle between the beam of light and the line through the searchlight perpendicular to the wall is pi/6
you are given with $\frac{d\theta}{dt}$ and a constant distance $x$..
you should make a relationship between $\theta, x$ and the other variable that i'll call $y$.. in fact, the relationship could be $\tan \theta = \frac{y}{x}$... do you get the picture?

Originally Posted by Mickey
Find the slope of the curve x^2 + xy + 2y^3 = 4 at the point (-2,1)
find $\frac{dy}{dx}$ and evaluate at that point..

4. Originally Posted by Mickey
Could anyone please help us out with these problems even just to show how to begin. It would be great and I would be very grateful. I have an exam coming up and I need to get these problems sorted.

Show that the volume of the solid obtained by rotating the part of circle of x^2 + y^2 = 1 that lies between x =0 and x = 1/2 about the x axis is (11pi)/24 cubic units.

Find the antiderivative between e and 1 of (4 ln(x) + 5) / x dx using substitution of w = 4ln(x) + 5

A searchlight rotates at a rate of 3 revolutions per min. Its beam of light hits a wall that is 10km away and produces a dot of light that moves horizontally along the wall. At what speed is the dot moving when the angle between the beam of light and the line through the searchlight perpendicular to the wall is pi/6

Find the slope of the curve x^2 + xy + 2y^3 = 4 at the point (-2,1)
For the first one it might be cooler to parameterize your curve. Let $x=\cos(t)$

and

let $y=\sin(t)$

Now we know that volume is given by

$\pi\int_a^{b}\left[y(t)\right]^2\cdot{x'(t)}dt$

So seeing that $\arccos(x)=t$
when $x=0\Rightarrow{t=\frac{\pi}{2}}$

and when

$x=\frac{1}{2}\Rightarrow{t=\frac{\pi}{3}}$

So we would have

$-\pi\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin^2(t)\c dot{-\sin(t)}dx=\pi\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sin^3(t)dt$

Now obviously

$\int\sin^3(t)dt=\int\sin(t)(1-\cos^2(t))dt=\int\bigg[\sin(t)-\sin(t)\cos^2(t)\bigg]dt=$ $-\cos(t)+\frac{\cos^3(t)}{3}$

So we have our volume is given by

$\pi\bigg[-\cos(t)+\frac{\cos(t)^3}{3}\bigg]\bigg|_{\frac{\pi}{3}}^{\frac{\pi}{2}}=\frac{11\pi }{24}\text{ cubic units }$
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For your second one I belive you have

$\int_1^{e}\frac{4+\ln(x)}{x}$

If we let as you suggested $w=4+\ln(x)\Rightarrow{dw=\frac{1}{x}}$

and then seeing that it is probably actually easier to not change the limits and just integrate this normall, sub back in and get our integral. So we would just compute

$\int{u\text{ }du}=\frac{u^2}{2}$

So subbing back in we get

$\int\frac{4+\ln(x)}{x}dx=\frac{(\ln(x)+4)^2}{2}$

Which implies that

$\int_1^e\frac{4+\ln(x)}{x}dx=\bigg[\frac{(\ln(x)+4)^2}{2}\bigg]\bigg|_{1}^e$
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For the last one we must use implicit differentiation.

So for
$x^2+xy+2y^3=4$

Differentiating, implicity mind you, we get

$2x+xy'+y+6y^2\cdot{y'}=0$

Now we need to find $y'$ at $(-2,1)$

So instead of solving for y' which is what you usually do when you are just asked to find the general formula, just plug in the numbers and solve for y' numerically

so we would have

$2(-2)+(-2)y'+1+6(1)^2y'=0\Rightarrow{-3+4y'=0}$

Solving gives us

$y'=\frac{3}{4}$

5. A searchlight rotates at a rate of 3 revolutions per min. Its beam of light hits a wall that is 10km away and produces a dot of light that moves horizontally along the wall. At what speed is the dot moving when the angle between the beam of light and the line through the searchlight perpendicular to the wall is pi/6

3 rev per min = $2{\pi}(3)=6{\pi} \;\ \frac{rad}{min}$

We want dx/dt given that $\frac{d{\theta}}{dt}=6{\pi}$

$x=10tan({\theta})$

$\frac{dx}{dt}=10sec^{2}({\theta})\frac{d{\theta}}{ dt}$

$\frac{dx}{dt}=10sec^{2}(\frac{\pi}{6})(6{\pi})=80{ \pi} \;\ \frac{km}{min}$