Question

I have a couple questions on a calculus worksheet I have.
What is the position of S(x) of a moving object if its velocity "v" is defined as v(x)=9x^2-4x+2 and S(0)=11?

Which of the following is the solution to the differential equation dy/dx=(x-5)/(y+1); y cannot equal -1
I need to be guided through, I don't just want the answer please. Thanks!

Answer 1

integrate
\( s(x) = \int v(x) \ dx \)

for the second , you separate the variables, put y and dy on one side, and the x's and dx on the other

Answer 2

I integrated and found the c to be 11, is that right? making the new equation \[3x^2-2x^2+2x+11\]
I didn't know why else they would give me an initial condition.

Why would I get them on two different sides? I thought the point was to have the dy/dx isolated?

Answer 3

yes, when x is 0, your equation gives 11

Answer 4

one technique to solve diff eq's is to separate the variables.
i.e. treat dy and dx as variables that we can separate

Answer 5

Okay, so the answer to the first question is just 3x^2-2x^2+2x+11?

So I tried to do what you did but I ended up looking at my page for about five minutes...

Answer 6

@phi

Answer 7

dy/dx=(x-5)/(y+1);
\[ \frac{dy}{dx} = \frac{(x-5)}{(y+1)} \\(y+1) \ dy = (x-5) \ dx
\]
integrate
\[ \int (y+1) \ dy = \int (x-5) \ dx \\
\frac{y^2}{2} + y = \frac{x^2}{2} -5x + c \]
which if we work at it, can be put in the standard form for the equation of a circle.

Answer 8

Okay, I understood everything you did all the way up until you started talking about circles. How would I change it to a circle if I have a +c, do I just tag it on to the end?

Answer 9

@phi

Answer 10

And shouldn't both antiderivatives have the +c?

Answer 11

oh, it's a hyperbola

As for the arbitrary constant, there are "properties" of arbitrary constants you might want to learn. For arbitrary a and b
2a --> a (the idea is 2a is still arbitrary... can be any value)
a+ 3 --> a (adding a constant is still arbitrary)
a+b= a --> (it is still some arbitrary number)

thus if we multiply through by 2, we get 2c , but rename it as c (still arbitrary)
\[ \frac{y^2}{2} + y = \frac{x^2}{2} -5x + c \\
y^2 +2y = x^2 -10x +c
\]
if we "complete the square"
https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/ex1-completing-the-square
we can write this as
\[ (y+1)^2 = (x-5)^2 + c \\ (y+1)^2-(x-5)^2 = c \]
and that is the equation of a hyperbola with center (5,-1)
It's properties will depend on the value of c, and we need some "initial condition" to determine c.

Answer 12

Any questions ?

Answer 13

One last question: A projectile is fired straight up from a platform 24 ft above the ground, with an initial velocity of 500ft/sec. Its downward acceleration is 32 ft/sec^2. What is the equation for the height of the projectile above the ground as a function of time "t" t=0 when the projectile is fired?

Answer 14

@phi

Answer 15

You can use calculus/physics
\[ V(t) = \int -32 \ dt \\ V(t)= -32 t+C \]
at t=0, the initial velocity is 500 ft/sec, so
V(t)= 500 - 32 t

next
\[ h(t) = \int V(t) \ dt = \int 500 - 32 t \ dt \\
h(t) = 500 t - 16t^2 + C \]
at t= 0, the height is 24, so C=24, and
\[ h(t)= 24 + 500 t - 16 t^2 \]

Answer 16

Oh, I got all the way to the position function, but then I didn't know what to do. So the height is 24 at t=0, well that makes sense now. Thanks!