Winter Assignment.. *Sigh* :oops:

[Coraelle]

Well, the first part of the assignment was to find a good calculus help site.. *ish here*... I googled calculus and the sites I visited so far are all either trying to sell me something, or look really amateur. Sparknotes has been good to me of course. But if anyone knows any other good sites, lemme know please

Okies.. second part of assignment:

Curve: y^2 = 2 + xy

I found that dy/dx = y/(2y-x)

Now.. I can't find:

a. All the points on the curve where slope = 1/2. I somehow got (0,1) as one of the points, dunno if that's correct.
b. Show that there are no points on curve where slope = 0
c. Let x and y be functions of time that are related by the equation y^2 = 2 + xy. At time t=5, the value of y is 3 and dy/dt=6. Find value of dx/dt.

Thanks I'm very very bad at implicit differentiation.. it drives me insane.

 

[Unco]

Okies.. second part of assignment:

Curve: y^2 = 2 + xy

I found that dy/dx = y/(2y-x)
Correct.

Now.. I can't find:

a. All the points on the curve where slope = 1/2. I somehow got (0,1) as one of the points, dunno if that's correct.

Set \(\displaystyle \mbox{\frac{dy}{dx} = \frac{1}{2}}\). Multiply both sides by \(\displaystyle \mbox{2y-x}\) (for \(\displaystyle \mbox{2y-x\neq0)}\). What must \(\displaystyle \mbox{x}\) equal?

Substitute your value for \(\displaystyle \mbox{x}\) into the curve's equation to find the corresponding y-ordinates.

b. Show that there are no points on curve where slope = 0

After completing (a.), what can you come up with for this?

c. Let x and y be functions of time that are related by the equation y^2 = 2 + xy. At time t=5, the value of y is 3 and dy/dt=6. Find value of dx/dt.

Apply the chain rule:
\(\displaystyle \L \mbox{\frac{dx}{dt} = \frac{dx}{dy} \cdot \frac{dy}{dt}}\)

Use the curve's equation to find \(\displaystyle \mbox{x}\) when \(\displaystyle \mbox{y=3}\).

 

[Coraelle]

Here's what I did...

a. y=(1/2)(2y-x) = y-(x/2). -x/2=0 so x= 0
and then substitute into derivative.. so point= (0,1)

b. I set y/(2y-x)=0 and I got y=0, so I put that into the derivative and I got 0/-x = 0 so... x can be any real number? Er..

c. 9=2+3x so, x= 7/3. And dx/dt=6 * dx/dy?
And um, dx/dy is dy/dx flipped around? dx/dy=(2y-x)/y
dx/dy = (12*3 - 6*(7/3))/3 = 22/3 = 7.33

thanks for replying

 

[Unco]

Re: Here's what I did...

a. y=(1/2)(2y-x) = y-(x/2). -x/2=0 so x= 0
and then substitute into derivative.. so point= (0,1)

Set dy/dx = 1/2:

\(\displaystyle \mbox{\frac{y}{2y - x} = \frac{1}{2} \Rightarrow y = y - \frac{1}{2}x}\) (for \(\displaystyle \mbox{2y - x \neq 0) \Rightarrow x = 0}\)

That is, the slope of the curve is 1/2 at x=0. Substitute x=0 into the curve's equation to find the corresponding y-ordinates.

b. I set y/(2y-x)=0 and I got y=0, so I put that into the derivative and I got 0/-x = 0 so... x can be any real number? Er..

y=0 (for \(\displaystyle \mbox{2y - x \neq 0\)) is correct. What do you notice when you substitute \(\displaystyle y = 0\) into the curve's equation, though?

There is no point on the curve where \(\displaystyle \mbox{y=0}\)!

c. 9=2+3x so, x= 7/3.
and..then..what do I do now?

Plug \(\displaystyle \mbox{x=\frac{7}{3}}\) and \(\displaystyle \mbox{y=3}\) into \(\displaystyle \mbox{\frac{dx}{dy} = \frac{2y - x}{y}}\) and apply the chain rule.

 

[Coraelle]

ok...

9=2+3x so, x= 7/3. And dx/dt=6 * dx/dy?
And um, dx/dy is dy/dx flipped around? dx/dy=(2y-x)/y
dx/dy = (12*3 - 6*(7/3))/3 = 22/3 = 7.33

I have two more problems, I just finished drawing th figures for them

 

[Coraelle]

Ok, first graph and problem: I accidentally messed up labeling the first graph, so where you see 1/(x^2) it is actually 1/ (x^ (1/2)). Anyways, the problem goes:

Line t is tangent to the graph of y= 1/ (x^ (1/2)) at point P, with coordinates (w,1/w^2) where w > 0. Point Q has coordinates (w,0). Line t crosses the x axis at the point R, with coordinates (k,0)

a. find the value of k when x = 3
b. for all w>0, find k in terms of w
c. Suppose that w is increasing at the constant rate of 7 units per second, when w = 5, what is the rate of change of k with respect to time?

[/img]

 

[Coraelle]

2nd graph

And for the second graph and question:
The graph of f’(x) us shown on the closed interval [-1,5]. The graph of f’ has horizontal tangent line at x = 1 and x = 3. The function f is twice differentiable with f(2)=6
a. Find the x-coordinate of each point of inflection of the graph of f. Give reason for your answer.
b. At what value of x does f attain its absolute minimum value and absolute maximun value on closed interval. Show analysis that leads to answer.
c. Let g be the function defined by g(x)=x * f(x). Find an equation for line tangent to the graph of g at x=2.

 

[Coraelle]

Okay

For the first problem, what threw me off is the "x=3" I thought the x was refering to the x in the function y= 1/(x^(1/2)). But, maybe its just refering to the coordinates, in that case x=w..

 

[Coraelle]

Yay! I figured out 1 a. and b. Please check though

a. When x=3, w=3 P3, 1/9)

y=x^(-1/2) y'= (1/2)x^(-3/2)

Slope = -1/2(3)^(-3/2)=-0.1

y-(1/9)=-0.1(x-3)

y=-0.1x+0.411 k=4.11

b. slope = (-1/2)w^(-3/2)
y-1/9 = (-1/2)w^(-3/2)x + (1/2)w^(-3/2) y=0

-1/9=(-1/2)w^(-3/2)x+(1/2)w^(-3/2)

-1/9=(1/2)w^(-3/2) * (-x+1)

-x+1 = 2/(9w^(-3/2))


k = - (2w^(3/2))/9 + 1

Did I do it right? I tried c, but I have no idea how to even start.. the rate of change of 7? Does that mean that the instantaneous rate of change, or y'(5)=7?

 

[Coraelle]

ooopss

I messed up b.. all the parts where you see 1/9 it should be 1/w^2

 

[Coraelle]

Continued

so, instead my answe is k=2/w^(1/2) + 1

 

[Unco]

Re: Okay

For the first problem, what threw me off is the "x=3" I thought the x was refering to the x in the function y= 1/(x^(1/2)). But, maybe its just refering to the coordinates, in that case x=w..
I think it only makes sense to say "when w=3".

a. When x=3, w=3 P (3, 1/9)

y=x^(-1/2) y'= (1/2)x^(-3/2)
y' = -(1/2)x^(-3/2)

Slope = -1/2(3)^(-3/2)=-0.1
It has the slope of the curve at the point (w, 1/w^2), so this equation has followed from: slope = (-1/2)w^(-3/2). And use more decimal places if you do not wish to use exact values (radicals and such).

y-(1/9)=-0.1(x-3)

y=-0.1x+0.411 k=4.11
There are some rounding issues here, but you have the idea.

b. slope = (-1/2)w^(-3/2)

The line through (w, 1/w^2) with slope (-1/2)w^(-3/2) has equation:
y - 1/w^2 = (-1/2)w^(-3/2) (x - w)

y - 1/w^2 = (-1/2)w^(-3/2)x + (1/2)w^(-3/2)w

y=0

-1/w^2 = (-1/2)w^(-3/2)x+(1/2)w^(-3/2)

-1/w^2 = (1/2)w^(-3/2) * (-x+1)

-x+1 = 2/(9w^(-3/2))

k = - (2w^(3/2))/9 + 1

That initial error got you there.

Start from the beginning:
y - 1/w^2 = (-1/2)w^(-3/2) (x - w)

Set y=0:

-1/w^2 = (-1/2)w^(-3/2) (x - w)

And just don't bother expanding. Divide both sides by (-1/2)w^(-3/2) and you're pretty much there.

(Although we might normally plug the coordinates (k, 0) in at the start and solve for k.)


Did I do it right? I tried c, but I have no idea how to even start.. the rate of change of 7? Does that mean that the instantaneous rate of change, or y'(5)=7?
"w is increasing at the constant rate of 7 units per second" means dw/dt = 7.

Apply the chain rule like your previous question.