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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,770

hope this is correct so far... up to d.

but why is there an introduction of t when so far we just have x and y?

also I assumed the $|-4|$

this is from a calc I hw..

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,770

hope this is correct so far... up to d.

but why is there an introduction of t when so far we just have x and y?

also I assumed the $|-4|$

this is from a calc I hw..

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- #2

b) Correct, although I would substitute for $y$ instead:

\(\displaystyle x=-2y\)

\(\displaystyle 4y^2-8y^2+y^2=-12\)

\(\displaystyle y^2=4\)

\(\displaystyle y=\pm2\)

c) Correct.

d) Correct. For clarity, I think I would use the chain rule:

\(\displaystyle \frac{dx}{dt}= \frac{dx}{dy}\cdot\frac{dy}{dt}= \left(-\frac{1}{4} \right)\left(-\frac{1}{2} \right)= \frac{1}{8}\)

$t$ is simply a parameter that has been introduced. Normally $t$ represents some unit of time. Another approach would be to begin with the curve:

\(\displaystyle x^2+4xy+y^2=-12\)

Differentiate with respect to $t$ then divide through by 2:

\(\displaystyle x\frac{dx}{dt}+2\left(x\frac{dy}{dt}+\frac{dx}{dt}y \right)+y\frac{dy}{dt}=0\)

Plug in the given data \(\displaystyle \left(x,y,\frac{dy}{dt} \right)=\left(-4,14,-\frac{1}{2} \right)\) to get:

\(\displaystyle -4\frac{dx}{dt}+4\left(1+7\frac{dx}{dt} \right)-7=0\)

\(\displaystyle 24\frac{dx}{dt}=3\)

\(\displaystyle \frac{dx}{dt}=\frac{1}{8}\)

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- #3

- Jan 31, 2012

- 2,770

nice to see a more condensed version of this

although I was encouraged that I got the answers correct.

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- #4

You did well, nothing wrong with your work at all. It is always nice to see someone post their actual work too.

nice to see a more condensed version of this

although I was encouraged that I got the answers correct.

In part b), since you want $y$-values, it is simply a bit more direct to substitute for $x$ so that you can solve for $y$ directly.