# [SOLVED]if correct what is t

#### karush

##### Well-known member
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

Staff member
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
well that was certainly very helpful
nice to see a more condensed version of this

although I was encouraged that I got the answers correct.

#### MarkFL

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.