My question is as follows; In the book Anton Calculus 12th edition, Question 56 in section 5.2 has a solution contradicting my own; I will break the reasoning down for my answer to aid a reader in helping me. Or, possibly the book's solution is incorrect.
Problem:
Find an equation of the curve that satisfies the given. At each point (x,y) on the curve, the slope equals three times the square of the distance between the point and the y-axis; the point (-1,2) is on the curve.
Book Solution: \begin{align*} dy/dx=x^2,y&=\int x^2dx = x^3/3+C;\\ y&=2 \textrm{ when } x=-1,\textrm{ so } (-1)^{3}/3+C = 2, C=7/3\\ \textrm{thus } y&=x^{3}/3+7/3 \end{align*} My solution:
Here is my first thought for solving this, possibly $\vec{v} = (x,y)$ then we get, $$\frac{dy}{dx} = 3\|\vec{v}||^{2}\implies y =\int3\|\vec{v}||^{2}dx $$ After doing said integration, solve the initial value problem to get $C_{1}$.
The above must be incorrect since this would mean that $\vec{v}$ is a vector from the origin; So I will do my best to break my thought process into an algorithm the reader can understand, which may aid in helping me.
What I think I know. The labels (1, 2, ... 7) can be used to aid in replying to my question.
- (1) F(x) = y
- (2) "at each point", every (x,F(x)).
- (3) "on the curve", the equation F(x).
- (4) "the slope", $\frac{dy}{dx}$
- (5) "equals three times the square," three times a square is 3(x^2).
- (6) "the distance between the point and the y-axis", Since it's asserting the distance between the point and the y-axis, I think it's safe to assume what ever vector I come up with will not be going through the origin, so possibly if $P1 = (0,y)$, and $P2 = (x,y)$ then $\vec{P1P2} = (x,y)-(0,y) = (x,y-y) = (x,0)$
- (7) "the point (-1,2) is on the curve". This will be used to find our $C_{1}$
Here is my Second thought. Using $\vec{P1P2}$ from (6) \begin{align} \frac{dy}{dx} &= 3||\vec{P1P2}||^2\\ &= 3x^2 \\ \therefore y &= \int3x^2dx = 3\frac{1}{3}x^{3}=x^3+C_{1} \\ \end{align} Finally finding $C_{1}$ (7) \begin{align} y(-1) & = (-1)^3 + C_{1}\\ \implies 2 &= -1 +C_{1} \\ 3 &= C_{1} \end{align}
So we now have, $F(x) = y = x^{3} + 3$.
Any help that can be provided is greatly appreciated; thanks in advance.
I have taught out of Anton and other similar calculus texts and can attest that it is not unusual to find exercises with incorrect 'answers' in the key at the back of the book.
When teaching from a new text, it did not take long to find many of these because students would ask me about those problems when they could not get the 'correct' answer. I would always mark these in my copy of the text so that I could forewarn students and sent a note to the publisher so that a correction could be made in the next addition.
Since OP's solution to the exercise is correct, this is obviously a mistake in the answer key.