Homework help...

[MalTalm]

Since we're on the topic, does anyone here have any experience with Simulink or LaTeX?

 

[David.]

1. Find the point(s), if any, on the graph of the function

48x^3

at which the tangent line is parallel to the line

9x-y+2=0

so do I put the equation in the second part into slope intercept form to get my slope? Then what?

2. find the value of dy/dx at the indicated point

y=1-4x+x^3 at (1,-2)

I move y to the other side, 0=1-4x+x^3-y, then find derivatives:

0=4+3x^2-1 then do I plug in the x value? to get.. 6? thats not the right answer. I think im missing a final step

3)

find value V of a car after T years is v(t)=11---/t+2000 t is greater than equal to 1 and less and equal to 6....
What is the instantaneous rate of change after 2 years?


i just plugged in 2 for t and solved, but that isnt right... its not the instanaeous rate of change..

 

[Kouki]

having graduated with a mechanical engineering degree about a year ago

i'm disappointed to say that I am of no help xd

wolfram alpha

 

[David.]

wolfram alpha hates these questions and tells me to fuck myself

 

[MalTalm]

1. Find the point(s), if any, on the graph of the function

48x^3

at which the tangent line is parallel to the line

9x-y+2=0

so do I put the equation in the second part into slope intercept form to get my slope? Then what?

2. find the value of dy/dx at the indicated point

y=1-4x+x^3 at (1,-2)

I move y to the other side, 0=1-4x+x^3-y, then find derivatives:

0=4+3x^2-1 then do I plug in the x value? to get.. 6? thats not the right answer. I think im missing a final step

3)

find value V of a car after T years is v(t)=11---/t+2000 t is greater than equal to 1 and less and equal to 6....
What is the instantaneous rate of change after 2 years?


i just plugged in 2 for t and solved, but that isnt right... its not the instanaeous rate of change..

These just have to do with understanding what derivative and integral are.

The derivative is the instantaneous slope of a function at any point. (Derivative = slope)
The integral is the area under the curve. That said:

the graph of 48x^3 ->
(1) y1=48x^3
the derivative of y in terms of x...
(2) dy1/dx=144x^2

The line you need to run parallel to is y2=9x+2
(3)dy2/dx=9

set dy2/dx from (3) = dy1/dx in (2)

We get
9=144x^2 -> 9/144=x^2 -> sqrt(9/144)=x
there are two unique solutions...
x=.25
and x=-.25

to find the points, we need corresponding y values. so we plug our solutions back into equation (1)

y= 48*(.25)^3 ->
y= .75
y= 48*(-.25)^3 ->
y= -.75
so our points are
(.25,.75)
and (-.25,-.75)

because our function goes through the origin and is an odd function, we should anticipate two solutions reflected about y=-x, which we have.

(2) This one is a bit less involved...
Remember, y is a dependent variable, x is the only independent variable. If we were presented with a parametric equation, we might need to solve for x and y separately. Alas, we were given a standard equation, so the y value is insignificant in our calculation...

y=1-4x+x^3 at (1,-2)
so
(1)dy/dx=4+3x^2
plug x=1 into equation 1
dy/dx=4+3*(1)^2

dy/dx=7

(3) not sure about your equation here. You might need to rewrite this...
v(t)=(11(---???)/t)+2000
I assume the 2000 is added to the quantity, and not added to t in the denominator, is that right?
also, what are the --- in the numerator supposed to represent? Let me know, and I'll help you out.

EDIT* - BTW Instantaneous rate of change is just another way of saying slope, so take the derivative of the function and plug in t. If you need help with that, just rewrite the equation. Good luck!

 

[David.]

my bad, hit --- instead of 000.. its 11000...

got the first two numbers for the 1st one you did, just needed to plug those in to get y values..

the second one still isnt correct apparently... I put 6 (because I didnt take the 1 out when finding the derivative), but 7 is also wrong. hm.

 

[MalTalm]

my bad, hit --- instead of 000.. its 11000...

got the first two numbers for the 1st one you did, just needed to plug those in to get y values..

the second one still isnt correct apparently... I put 6 (because I didnt take the 1 out when finding the derivative), but 7 is also wrong. hm.

yup, i forgot to track the - sign on the second problem...

dy/dx=-4+3x^2
so dy/dx=-1

 

[David.]

noticed it too.. got it on the last try (phew)

 

[MalTalm]

my bad, hit --- instead of 000.. its 11000...

got the first two numbers for the 1st one you did, just needed to plug those in to get y values..

the second one still isnt correct apparently... I put 6 (because I didnt take the 1 out when finding the derivative), but 7 is also wrong. hm.

so the equation is then
(1) v(t)=(11000/t)+2000, yes?

if so,
(2) dv(t)/dt = -11000/t^2
plug t=2 into equation 2 to get instantaneous rate of change...

dv(t)/dt= -11000/4
dv(t)/dt= -2750

 

[Walter White]

I don't believe in math. Its not real.

 

[David.]

so finding rate of change (which is slope) is basically predicated on finding the derivative and then plugging in the x values?

I had one more that I thought i could figure out by seeing the answer to that last one, but I was wrong..

p=(20000/7x+120) -7, x is greater than 0, less than 100..

find rate of change of price with respect to x.. so I find the derivative (unless im wrong, is (20000/7)-7 but I have nowhere to plug in an x value..

 

[DJTJ]

Statistics two, really easy. One way anovas and one/two porportions using minitab. However, how the fuck do you find degrees of freedom, mean square, mean square error, and f value?

Source SS DF MS F
Treatments SST k-1 SST / (k-1) MST/MSE
Error SSE N-k SSE / (N-k)
Total (corrected) SS N-1

that's the chart of equations, but it makes no fucking sense. need to knock this test out, and my teacher can't teach.

 

[MalTalm]

so finding rate of change (which is slope) is basically predicated on finding the derivative and then plugging in the x values?

I had one more that I thought i could figure out by seeing the answer to that last one, but I was wrong..

p=(20000/7x+120) -7, x is greater than 0, less than 100..

find rate of change of price with respect to x.. so I find the derivative (unless im wrong, is (20000/7)-7 but I have nowhere to plug in an x value..

It's sometimes useful to rewrite the denominator into the numerator, so the equation would become

p=(20000*(7x+120)^-1) -7
dp/dx= (20000*(-1)(7x+120)^-2)*7
or, simplified:
dp/dx= -140000/(7x+120)^2

This is using the chain rule, which states d/dx of (ax+b)^c is the derivative of the outside term multiplied by the derivative of the inside term.

It's a pain to get used to at first, but it will become on of the most common derivative rules you'll apply in calculus.

 

[Surge]

I was great at math until they started putting letters in there.

 

[Ohio]

I was great at math until they started putting letters in there.

:chuckles: