MHB Solve $|-4|$: What's the Answer?

[karush]

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]

a) Correct.

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

$$x=-2y$$

$$4y^2-8y^2+y^2=-12$$

$$y^2=4$$

$$y=\pm2$$

c) Correct.

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

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

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

$$x^2+4xy+y^2=-12$$

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

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

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

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

$$24\frac{dx}{dt}=3$$

$$\frac{dx}{dt}=\frac{1}{8}$$

 

[karush]

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.

 

[MarkFL]

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.

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.