Question

AP Calculus: Slope Fields/ Separate Variables
Find the equation of the curve, given the derivative and the indicated point on the curve.
dy/dx = -1/x^2 dx

Answer 1

given point on graph is (1,3)

Answer 2

@SithsAndGiggles

Answer 3

Is that
\[\frac{dy}{dx}=-\frac{1}{x^2}?\]

Answer 4

yeah

Answer 5

\[dy=-\frac{dx}{x^2}\\
\int dy=-\int \frac{dx}{x^2}\]

Answer 6

so how would i integrate that

Answer 7

Well, you can recognize that \(\displaystyle-\frac{1}{x^2}\) is the derivative of \(\displaystyle\frac{1}{x},\) or you can write the RHS as
\[-\int x^{-2}dx,\]
then apply the power rule for integrating.
Recall that the power rule is
\[\int x^ndx=\frac{x^{n+1}}{n+1}+C\]

Answer 8

ok so the answer is \[y=\frac{ -1 }{ x^2 } + 3\]

Answer 9

Not quite. After integrating, you get
\[y=\frac{1}{x}+C\]

Answer 10

Because the derivative of this (in differential form) is
\[dy=-\frac{1}{x^2}dx+0\\
dy=-\frac{1}{x^2}dx\]

Answer 11

so that's the answer? really? i'm so confused :S

Answer 12

Which part are you confused about?

Answer 13

that the answer isn't \[y=\frac{ -1 }{ x^2 } + 3\]

Answer 14

the given point on graph is (1,3)

Answer 15

I'll go through the solution-finding step-by-step:

You're given
\[\frac{dy}{dx}=-\frac{1}{x^2}\]
Move the dx to the right:
\[dy=-\frac{1}{x^2}dx\]
Integrate both sides:
\[\int dy=\int\left(-\frac{1}{x^2}\right)dx\]
I'll use the power rule here; I rewrite 1/x² as x^(-2):
\[\int dy=\int\left(-x^{-2}\right)dx\\
y=-\left(\frac{x^{-2+1}}{-2+1}\right)+C\\
y=-\left(\frac{x^{-1}}{-1}\right)+C\\
y=-\left(-\frac{1}{x}\right)+C\\
\color{red}{y=\frac{1}{x}+C}\]
Given the point (1,3), you have
\[3=\frac{1}{1}+C\\
3=1+C\\
2=C\]
So, the solution is
\[y=\frac{1}{x}+2\]
The red is where the misunderstanding seems to take place.

Answer 16

yeah that's what i put i think i forgot to subtract the 1 to the other side lol

Answer 17

oh wait nvm

Answer 18

ok ty :) can you help me with another one please? ^^

Answer 19

yep