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[SOLVED] if correct what is t

karush

Well-known member
Jan 31, 2012
2,770
View attachment 1760
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..
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
a) Correct.

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:

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

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

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

karush

Well-known member
Jan 31, 2012
2,770
well that was certainly very helpful (Happy)
nice to see a more condensed version of this

although I was encouraged that I got the answers correct.:cool:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
well that was certainly very helpful (Happy)
nice to see a more condensed version of this

although I was encouraged that I got the answers correct.:cool:
You did well, nothing wrong with your work at all. It is always nice to see someone post their actual work too. (Yes)

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.