differentiate dx/dt=kt for k?

Question

Hi. The question asks:
"If dx/dt=kt, and if x=2 when t=0 and x=6 when t=1, then k="
After differetiaing there are 2 variables, k & c which confuses me. The possible answers are:
a)ln4
b)8
c)e^3
d)3
e)none
I believe the answer is 8 but I don't know how to get there.
Thanks for the help,
Amit

Also:
There is another problem which looks simple but I think I'm missing something. It's:
"If dy/dx=e^y and y=0 when x=1, then..."
Possible answers:
a)y=ln|x|
b)y=ln|2-x|
c)e^(-y)=2-x
d)y=-ln|x|
e)e^(-y)=x-2
Thanks. Amit.

Answer 1

dx/dt=kt
dx=ktdt
take the integral both sides
x=kt^2/2 +C
if t=0, x=2
plugging that values into x=kt^2/2+C
2=k(0)^2/2+C
therefore C=2
in the general formula is x=kt^2/2+2
if t=1 , x=6 then k=?
6=k(1)^2/2+2
6=k/2+2
4=k/2
k=8 which is B
b.There is another problem which looks simple but I think I'm missing something. It's:
"If dy/dx=e^y and y=0 when x=1, then..."
Possible answers:

so dy/e^y=dx
take integral
-1/e^y=x+C
y=0, x=1
therefore, -1/e^0=1+C
-1/1=1+c
-1=1+c
c=-2
therefore
-1/e^y=x-2
-e^-y=x-2
e^-y=2-x
which is C

Answer 2

OK, here's how to do the first one (they're both almost the same): Multiply through by dt so that all the t's are on one side and all the x's on the other (move all the variables to opposite sides):
dx = kt dt
Now integrate both sides. The lower bound on the left will be x=2 while the lower bound on the right is t=0 (the lower bound is the starting point). The upper bound on the left is x=6 while the upper bound on the right is t=1.
Now you have an equation for k


If dy/dx=e^y and y=0 when x=1, solve for y in terms of x:
Use the same procedure.
Multiply through by dx:
dy = e^y dx
Uh oh. There's both x and y on the right. Better move the y's over to the other side by dividing through by e^y:
e^-y dy = dx.

Now we are ready to integrate both sides. The lower bounds are y=0 on the left and x=1 on the right. But what about the upper bounds? Use y=y on the left and x=x on the right.

One really nice thing about this technique is that you don't have to worry about always remembering to include that constant C for your integration. By matching up the correct bounds, the constant is automatically incorporated into the result, no muss, no fuss.