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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Gloamglozer on April 26, 2010, 07:43:50 pm

Title: Implicit Differentiation & Related Rates
Post by: Gloamglozer on April 26, 2010, 07:43:50 pm: 1. We have the equation $x^4 + y^4 = 3^4$ . Find the points on the curve which the tangent line is parallel to $y=x$ .

I've differentiated it and got:

$\frac{d}{dx}(x^4) + \frac{d}{dy}(y^4)\frac{dy}{dx} = \frac{d}{dx}(3^4)$

$4x^3 + 4y^3\frac{dy}{dx} = 0$

$4y^3\frac{dy}{dx} = -4x^3$

$\frac{dy}{dx} = \frac{-4x^3}{4y^3}$

$\frac{dy}{dx} = \frac{-x^3}{y^3}$

If it is to be parallel to $y=x$ , then that means that $\frac{dy}{dx}=1$ .

So the next thing I should do would be:

$\frac{-x^3}{y^3} = 1$ Right?

2. Just to note, this is an assignment question. Feel free to help me as much or as less as you wish.

I think the main problem is identifying the rates.

A street light is mounted at the top of a 5.4 metre pole. A man 1.8 m tall walks away from the light in a straight path at a speed of 1 m/s. How fast is the top of the his shadow moving when he is 10 m from the pole?

I've drawn a picture:
(http://img21.imageshack.us/img21/6295/polequestion.jpg)

From that, I've calculated the length of the shadow, which I can't find how it would relate to the question:

Let x be the length of the shadow.
$\frac{1.8}{x} = \frac{5.4}{10+x}$

$5.4x = 18 + 1.8x$

$3.6x = 18$

$x = 5$

I think what I need to do next is $\frac{Distance}{Time}$ to get the m/s but I can't figure out what to do next with setting up the chain rule.
Title: Re: Implicit Differentiation & Related Rates
Post by: Martoman on April 26, 2010, 07:49:35 pm: For the first question, I see no problems in what you have done.
Title: Re: Implicit Differentiation & Related Rates
Post by: m@tty on April 26, 2010, 07:49:55 pm: For 1. yeah, let $\frac{-x^3}{y^3}=1$ then $-x^3=y^3$ and $y=-x$ .

Sub that into the original relation, and solve.
Title: Re: Implicit Differentiation & Related Rates
Post by: m@tty on April 26, 2010, 08:41:28 pm: 2.

You want to find the rate of change of length of shadow with respect to time.

Let distance from man to pole = d. As he is walking at 1 m/s, d=t, where t is time in seconds.

Let length of shadow = l.

So you want $\frac{dl}{dt}$ which by the chain rule is equivalent to $\frac{dl}{dd}\cdot \frac{dd}{dt}$ .

You should be able to figure it out from there, keeping in mind your earlier workings...

I did the working then remembered it is an assignment, and you may not want me to do it for you... But it's here anyway, because I didn't want to waste typing it all out...

\frac{dd}{dt}=1 So \frac{dl}{dt}=\frac{dl}{dd}

From the similar triangles you set up \frac{1.8}{l}=\frac{5.4}{d+l}

1.8d+1.8l=5.4l

l=\frac{1.8}{3.6}d

\frac{dl}{dd}=\frac{1}{2}.

Therefore, from earlier working, \frac{dl}{dt}=\frac{1}{2} \cdot 1 ms^{-1}
Title: Re: Implicit Differentiation & Related Rates
Post by: Chavi on April 26, 2010, 08:51:19 pm: Question 2 involves a few steps:
a) let the distance between the dude and the pole = d (not 10, as this figure is subject to change)
b) express x in terms of d: $\frac{3.6}{d} = \frac{1.8}{x}$ (similar triangles) solving for x gives $x= \frac{d}{2}$
c) find the speed at which the length of the shadow is increasing:

$\frac{dx}{dt} = \frac{dx}{dd} \frac{dd}{dt}$

from before $x= \frac{d}{2}$ so $\frac{dx}{dd} = \frac{1}{2}$

$\Longrightarrow \frac{dx}{dt} = \frac{1}{2}$ as $\frac{dd}{dt} = 1m/s$

d) Express, in terms of x, the distance fo the end point, P, of the shadow

$|OP| = d + x = (2x) = x = 3x$

and thus the change in shadow size/movement with respect to time:

$\frac{d|OP|}{dt} = \frac{d|OP|}{dx} \frac{dx}{dt}<br />= 3 * \frac{1}{2}<br />= \frac{3}{2} m/s$

hope that helps
m@tty beat me to it :)
Title: Re: Implicit Differentiation & Related Rates
Post by: m@tty on April 26, 2010, 08:56:05 pm: Ah, yes. You have to add the speed of movement of the guy relative to the pole as well. So my final answer is not quite correct.
Title: Re: Implicit Differentiation & Related Rates
Post by: Ilovemathsmeth on April 27, 2010, 12:58:29 pm: Yep I got x = 3/2^0.25 and y = -3/2^0.25

I also got x = -3/2^0.25 and y = 3/2^0.25

I got 3/2 m/sec for the second part.
Title: Re: Implicit Differentiation & Related Rates
Post by: Gloamglozer on April 27, 2010, 01:43:36 pm: Quote from: Ilovemathsmeth on April 27, 2010, 12:58:29 pm
Yep I got x = 3/2^0.25 and y = -3/2^0.25

I also got x = -3/2^0.25 and y = 3/2^0.25

I got 3/2 m/sec for the second part.

Awesomeness. Good work ILLM. H1 on the way for you. :D
Title: Re: Implicit Differentiation & Related Rates
Post by: Ilovemathsmeth on April 30, 2010, 09:54:48 pm: I really hope so because my other subjects are going downhill... :S Good luck to you too!