derivative and tangent line problem

Ok, I'm doing a packet for math on derivatives and extrema and this is a problem where I honestly have no idea to start. I know what a curve, derivative, extrema, and tangent line together, but in this problem I don't know how to link them all together.

(noncalculator) the curve with derivative $\displaystyle (dy)/(dx) = (3-x)/(y+2)$ has y = -3 as a tangent line.

at what point is the line tangent to the curve? determine if the point that you found is a relative maximum point, relative minimum point or neither for the curve. Justify your answer.

Any help is appreciated. Thanks!!

Re: derivative and tangent line problem

If y=-3 then it looks like a stationary point, i.e where y'=0, that will help.

Re: derivative and tangent line problem

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Originally Posted by

**pickslides** If y=-3 then it looks like a stationary point, i.e where y'=0, that will help.

So are you saying where y = -3 on the original function is where y'=0 on the derivative function? if this is true, then that means I have to find out what x is when the derivative is equal to zero, but how do I do this since there is a y in the derivative equation?

Re: derivative and tangent line problem

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Originally Posted by

**TacticalPro** So are you saying where y = -3 on the original function is where y'=0 on the derivative function? if this is true, then that means I have to find out what x is when the derivative is equal to zero, but how do I do this since there is a y in the derivative equation?

Didn't you just say that y = -3?

Re: derivative and tangent line problem

As mentioned, the slope of the line y = -3 is zero, thus the slope of the function described by the differential equation will be zero when y = -3. So set your derivative equal to zero and let y = -3 and solve for x. (As you will see, you really don't need to know what y is...)

Then, to determine the nature of the function's behavior at that point, use the second derivative test, using the quotient and chain rules, and substitute into your formula for y'' the values for x, y, and y' to find the sign of y'' at the tangent point.

Re: derivative and tangent line problem

Quote:

Originally Posted by

**MarkFL2** As mentioned, the slope of the line y = -3 is zero, thus the slope of the function described by the differential equation will be zero when y = -3. So set your derivative equal to zero and let y = -3 and solve for x. (As you will see, you really don't need to know what y is...)

Then, to determine the nature of the function's behavior at that point, use the second derivative test, using the quotient and chain rules, and substitute into your formula for y'' the values for x, y, and y' to find the sign of y'' at the tangent point.

Ok, plugging -3 in for y I get $\displaystyle (3-x)/(-1)=0$ I then multiply both sides by -1 and do some further algebra to get x=-3.

Assuming this is correct, I then plug in x = -3 to the second derivative, which I began to evaluate but then ran into some complications

1. there is a y in the first derivative, does this mean I must do implicit differentiation?

2. Do I have to use chain rule for (3-x) and (y+2) even though they're not raised to any power?

Thanks

Re: derivative and tangent line problem

Quote:

Originally Posted by

**TacticalPro** Ok, plugging -3 in for y I get $\displaystyle (3-x)/(-1)=0$ I then multiply both sides by -1 and do some further algebra to get x=-3.

Assuming this is correct, I then plug in x = -3 to the second derivative, which I began to evaluate but then ran into some complications

1. there is a y in the first derivative, does this mean I must do implicit differentiation?

2. Do I have to use chain rule for (3-x) and (y+2) even though they're not raised to any power?

Thanks

I think you'll find that x = 3, not -3.

Re: derivative and tangent line problem

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Originally Posted by

**Prove It** I think you'll find that x = 3, not -3.

Oh yeah true that haha why do I always fail on basic math. Thanks for catching that

Re: derivative and tangent line problem

Quote:

Originally Posted by

**TacticalPro** Ok, plugging -3 in for y I get $\displaystyle (3-x)/(-1)=0$...

1. there is a y in the first derivative, does this mean I must do implicit differentiation?

2. Do I have to use chain rule for (3-x) and (y+2) even though they're not raised to any power?

Thanks

Yes, the implicit differentiation is what I meant by the chain rule, because that is what you use when dealing with y.

Re: derivative and tangent line problem

Quote:

Originally Posted by

**MarkFL2** Yes, the implicit differentiation is what I meant by the chain rule, because that is what you use when dealing with y.

oh so is that a yes to both questions? I didn't use chain rule, but I did use implicit differentiation and I got the second derivative to be $\displaystyle (-y^3-2y^2-y-2)/(-xy^2-4xy-4x+3)$

Is this correct? If not could you help walk me through? THanks

Re: derivative and tangent line problem

We have:

$\displaystyle \frac{dy}{dx}=\frac{3-x}{y+2}$

Differentiation with respect to x yields:

$\displaystyle \frac{d^2y}{dx^2}=\frac{(y+2)(-1)-(3-x)\frac{dy}{dx}}{(y+2)^2}$

Now use x = 3, y = -3 and dy/dx = 0...