Author Topic: Differentiation Help (Read 719 times) Tweet Share

johnsmith123 · « **on:** May 30, 2014, 09:45:03 pm »

Can someone please help with parts c & d? Thanks.

b^3 · « **Reply #1 on:** May 30, 2014, 10:38:27 pm »

$\begin{alignedat}{1}y & =2e^{\frac{x}{10}}+5 \\ \frac{dy}{dx} & =\frac{1}{5}e^{\frac{x}{10}} \\ \frac{dy}{dt} & =\frac{dy}{dx}\cdot\frac{dx}{dt} \\ & =\frac{1}{5}e^{\frac{x}{10}}\times-6 \\ \implies\frac{dy}{dt} & =-\frac{6}{5}e^{\frac{x}{10}} \end{alignedat}$
Part c:
Hint: integrate $\frac{dx}{dt}$ with respect to $t$ to get $x$ in terms of $t$ .

Spoiler

$\begin{alignedat}{1}\frac{dx}{dt} & =-6 \\ \implies x & =\int\left(-6\right)dt \\ & =-6t+C,\: C\in\mathbb{R} \\ x=0,\: t=0 \\ 0 & =0+C \\ \implies C & =0 \\ \implies x(t) & =-6t \\ \frac{dy}{dt} & =-\frac{6}{5}e^{\frac{-6t}{10}} \\ & =-\frac{6}{5}e^{\frac{-3}{5}t} \end{alignedat}$

Part D:
Hint: The speed will be given by the square root of the sum of the squares of the speed in the $x$ and $y$ directions (think Pythagoras), where each of these speeds is given by the rate of change of the distance over time (the derivative in that direction with respect to time).

Spoiler

$\begin{alignedat}{1}\frac{ds}{dt} & =\sqrt{\left(\frac{dy}{dt}\right)^{2}+\left(\frac{dx}{dy}\right)^{2}} \\ & =\sqrt{\left(-\frac{6}{5}e^{\frac{-3}{5}t}\right)^{2}+\left(-6\right)^{2}} \\ & =\sqrt{\frac{36}{25}e^{-\frac{6}{5}t}+36} \\ & =\sqrt{\frac{36}{25}\left(e^{-\frac{6}{5}t}+25\right)} \\ & =\frac{6}{5}\sqrt{e^{-\frac{6}{5}t}+25}\text{ units/second} \end{alignedat}$

Bestie · « **Reply #2 on:** May 31, 2014, 03:00:09 pm »

this is really random but...
can you please explain how you got the ds/dt formula, I know you said think pythagoras, but i still didn't really get it. sorry

how would you use that formula if you only had dx/dt, would it be?
ds/dt = squareroot [(dx/dt)^2 + (dy/dx)^2]?

thank you

b^3 · « **Reply #3 on:** May 31, 2014, 03:25:44 pm »

We should first note that we're dealing with a particle travelling in two dimensions. We're given orthogonal (well lets just call them perpendicular for the time being) components of the particles motion. That is the speed that it travels at in the $x$ direciton is $\frac{dx}{dt}$ and in the $y$ direction, $\frac{dy}{dt}$ . Since the particle is not solely travelling in the $x$ or $y$ directions (but rather travelling along a combination of the two), the actual speed of the particle (which we will call $\frac{ds}{dt}$ will be dependent on the components of speed.

Combining these two components to find the magnitude of $\frac{ds}{dt}$ gives $\frac{ds}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}$ .

Now sometimes you'll only be given a particle moving in one dimensions (such as in methods), where your speed would just be that of the speed of the particle moving in that direction, $\frac{dx}{dt}$ . You may also have a particle in three dimensions when you hit vector calculus, where the magnitude of the speed of the particle is the square root of the sum of the square of the rate of change of distance of each component with respect to time.

So if you only have $\frac{dx}{dt}$ and your particle is limited to a single dimension, then your speed is just $\frac{dx}{dt}$ , if you have the particle in two dimensions then you need to do what was done in the question above.

johnsmith123 · « **Reply #4 on:** June 01, 2014, 12:20:06 pm »

Thanks b^3 great explanation really appreciate it

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