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!