AP Calc: f(1)=4, f'(x) = 3x^2+1/2y for all (x,y)

[venialove]

Let f be a function with f(1) = 4 such that for all points (x,y) on the graph of f the slope is given by 3x^2+1 /2y.

a) Find the slope of the graph of f at the point where x=1.

b) Where an equation for the line tangent to the graph of f at x=1 and use it to approximate f(1,2)

c) Find f(x) by solving the inseparable differential equation dy/dx=3x^2+1/2y with the initial condition f(1)=4.

This is what i did:
y-y1=m(x-x1)
y-2=m(x-1)
y-2=4 (x-1)

 

[Deleted member 4993]

Let f be a function with f(1) = 4 such that for all points (x,y) on the graph of f the slope is given by 3x^2+1 /2y.

a) Find the slope of the graph of f at the point where x=1.
b) Where an equation for the line tangent to the graph of f at x=1 and use it to approximate f(1,2)
c) Find f(x) by solving the inseparable differential equation dy/dx=3x^2+1/2y with the initial condition f(1)=4.

This is what i did:
y-y1=m(x-x1)
y-2=m(x-1)
y-2=4 (x-1)

You probably meant

dy/dx = (3x^2 + 1)/(2y)

the way you wrote - it comes to be

dy/dx = 3x^2 + (1/2) * y

which one is correct? (You ought to know "order of operations" PEMDAS - from your elementary classes - and follow it)

In either case, the problem statement tells you to solve the differential equation first - please do that and show your work.

 

[venialove]

f(x)= 3x²+1/(2y)
f'(X)= 6x
F(1)=4
y-y1=m(x-x1)
y-1=6(x-4)
y-1=6x-24
y=6x-25

 

[stapel]

f(x)= 3x²+1/(2y)

So, in reply to the tutor's query, you're saying that the graph of the slope of the function is as follows?

. . . . .\(\displaystyle f(x)\, =\, 3x^2\, +\, \frac{1}{2y}\)

Since the above is defined as "f(x)", does this mean that the y is some constant, rather than a variable?

f'(X)= 6x
F(1)=4

How do "X" and "F" relate to the original "x" and "f"?

y-y1=m(x-x1)
y-1=6(x-4)
y-1=6x-24
y=6x-25

For which part of the exercise is this your solution? Have you yet attempted to solve the differential equation (which, as the tutor pointed out, is the first step)?

Please reply with clarification, including all of your work and reasoning. Thank you!

Eliz.

 

[skeeter]

Find f(x) by solving the inseparable differential equation dy/dx=3x^2+1/2y with the initial condition f(1)=4

well, it did say that the differential equation was inseperable, so I guess it can be narrowed down to either ...

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

or

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

{cue up "final Jeopardy" tune} :|