# Tutor profile: Taylor W.

## Questions

### Subject: Calculus

Solve the following integral: $(\int 4x \cos{(2-3x)}dx$)

This integral is easily solved with integration by parts. Recall integration by parts tells us that $(\int udv = uv - \int vdu$)A natural choice for $$u$$ and $$dv$$ are $$u = 4x$$ and $$dv = \cos{(2-3x)}dx$$. Next we solve for $$du$$ and $$v$$. $(u = 4x \rightarrow du = 4dx$)$(dv = \cos{(2-3x)}dx \rightarrow v = -\frac{1}{3}\sin{(2-3x)}$)By plugging into our integration by parts formula, we get $(\int 4x \cos{(2-3x)}dx = 4x(-\frac{1}{3}\sin{(2-3x)}) - \int-\frac{4}{3}\sin{(2-3x)}dx$) $( = -\frac{4}{3}x\sin{(2 - 3x)} + \frac{4}{9}cos{(2-3x)} + c$)With $$c$$ being the constant of integration.

### Subject: Algebra

Solve the following system of equations: $(\sin{x} + \cos{y} = 1.5\\ \sin^2{x} + \cos^2{y} = 1.25$)

A simple way to solve this trigonometric system of equations is to use the following substitution: $(u = \sin{x}\\ v = \cos{y}$)Using this substitution, our system of equations becomes$(u + v = 1.5\\ u^2 + v^2 = 1.25$)Our first equation yields $$ u = 1.5 - v $$. Substituting into our second equation we get $((1.5-v)^2 + v^2 = 4$)Which simplifies to $2v^2 - 3v + 1 = 0$)This quadratic has two roots, $$v_1 = 1$$ and $$ v_2 = 0.5 $$. Returning to our equation $$ u = 1.5-v$$ we get $(u_1 = 1.5-v_1 = 0.5$)$(u_2 = 1.5 - v_2 = 1$) As we near the end we need to solve four last equations:$(1) \sin{x} = 0.5, \cos{y} = 1$)$(2) \sin{x} = 1, \cos{y} = 0.5$) From Eq $$1)$$: $(\sin{x} = 0.5 \rightarrow x = (-1)^m\frac{\pi}{6} + \pi m, m \in Z$)$(\cos{y} = 1 \rightarrow y =2\pi l, l \in Z$) From Eq $$2)$$: $(\sin{x} = 1 \rightarrow x = \frac{\pi}{2} + 2\pi k, k \in Z$)$(\cos{y} = 0.5 \rightarrow y = \pm\frac{\pi}{3} + 2\pi n, n \in Z$) Which are our solutions to the system of equations.

### Subject: Physics

There is a notorious jewel thief speeding down the highway at velocity $$\vec{v}$$. A police officer, initially at rest, notices the pearl pilferer heading in his direction at a distance $$d$$. Assuming the officer cannot react in time to stop the thief when the thief initially passes the officer, at what rate must the officer accelerate to catch the thief makes it to a bridge a distance $$b$$ from the officer?

The only equation we will need for this problem is: $( \Delta x = \frac{1}{2}at^2 + v_i t$) where $$ \Delta x$$ is displacement, $$ a $$ is acceleration, $$ t $$ is time passed, and $$ v_i $$ is initial velocity. To begin, we first calculate the $$ maximum$$ amount of time the officer will have to catch the crook. To do this, we calculate the time it will take the burglar to reach the bridge. At a distance $$ d + b $$ and traveling at velocity $$ \vec{v} $$, with no acceleration, we can use our equation and find after a little algebra: $( t = \frac{d + b}{v} $)This is the most time that the officer can take to reach the burglar before he escapes across the bridge. We can now use this time to calculate the rate the officer must accelerate to travel a distance $$ b $$. We can plug in our value for $$t$$ above, $$ v_i = 0 $$ since the officer is initially at rest, and $$ \Delta x = b $$. Our equation becomes: $(b = \frac{1}{2}a (\frac{d + b}{v})^2 + 0$)Solving for $$a$$ we get: $(a = 2\frac{bv^2}{(d + b)^2}$) Which is our final answer.

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