# tricky calculus problem

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• Apr 23rd 2014, 12:11 AM
catenary
tricky calculus problem
Here's a problem from a calculus book that has me stumped:
Attachment 30739
I don't understand a single thing from the suggestion that the book is giving.. I know how to differentiate, I know how to solve for the equation of a line tangent at a point of a curve using point-slope formula, I know that the line that is normal to said line would have a reciprocal slope to the said line. but I don't know what the heck how to deal with this point Q(a,a^2) thing and everything mentioned in the suggestion. How do I attack this problem? Is this easy?
• Apr 23rd 2014, 05:19 AM
earboth
Re: tricky calculus problem
Quote:

Originally Posted by catenary
Here's a problem from a calculus book that has me stumped:
Attachment 30739
I don't understand a single thing from the suggestion that the book is giving.. I know how to differentiate, I know how to solve for the equation of a line tangent at a point of a curve using point-slope formula, I know that the line that is normal to said line would have a reciprocal slope to the said line. but I don't know what the heck how to deal with this point Q(a,a^2) thing and everything mentioned in the suggestion. How do I attack this problem? Is this easy?

Hello,

you are looking for a point which produces together with point P a straight line which is perpendicular to the graph of your function. That means:

1. Q denotes the point in Question.
2. Q lies on the graph of $\displaystyle p:y=x^2$. So if $\displaystyle x_Q = a$ then $\displaystyle y_Q = a^2$. Therefore $\displaystyle Q(a, a^2)$.
3. The derivative of p is $\displaystyle y'=2x$. Therefore in point Q the graph of p has the slope m = 2a .
4. That means: The perpendicular direction is described by: $\displaystyle m'=-\frac1{2a}$
5. The straight line QP has the equation (use 2 point formula of a straight line):
$\displaystyle QP:\frac{y-0}{x-18} = \frac{a^2-0}{a-18}~\implies~\boxed{y=\frac{a^2}{a-18} \cdot x - \frac{18a^2}{a-18}}$
6. That means: This straight line is perpendicular to the graph of the function p. Therefore: The slope of the line PQ must have the same direction as m':

$\displaystyle \frac{a^2}{a-18} = -\frac1{2a}$

Solve for a! As suggested by inspection. You should find an positive integer number.
• Apr 23rd 2014, 09:10 AM
catenary
Re: tricky calculus problem
big thanks earboth! I figured out points 1, 2, 3, and 4.. but I was writing the equation of the line QP as: y-0 = m(x-18), using the point-slope formula and i was substituting m'=-1/2a into m. So I actually had the equation of the line all I was missing was the slope... What I was missing was the other formula for m that you have reminded me of which was (y2-y1)/(x2-x1).. which when equated with the derived m of the line let's us solve for a. But this still doesn't explain what the book was talking about in it's "suggestion"... anyone here who gets it?
• Apr 23rd 2014, 10:58 AM
earboth
Re: tricky calculus problem
Quote:

Originally Posted by catenary
big thanks earboth! I figured out points 1, 2, 3, and 4.. but I was writing the equation of the line QP as: y-0 = m(x-18), using the point-slope formula and i was substituting m'=-1/2a into m. So I actually had the equation of the line all I was missing was the slope... What I was missing was the other formula for m that you have reminded me of which was (y2-y1)/(x2-x1).. which when equated with the derived m of the line let's us solve for a. But this still doesn't explain what the book was talking about in it's "suggestion"... anyone here who gets it?

Hello,

take the last equation and transform it a little bit:
$\displaystyle \frac{a^2}{a-18} = -\frac1{2a}~\implies~2a^3 = -a+18~\implies~2a^3+a-18=0$

This is the cubic equation which was mentioned in your book. You are aksed to find a small integer number for a:

a = 1, that means: $\displaystyle 2\cdot 1^3+1-18 =0$ is definitely false.
a=2, thast means: $\displaystyle 2\cdot 2^3+2-18 =0$ is definitely true. So a = 2 is a solution of this equation.

Therefore you know now that the LHS of the equation contains the factor (a-2). You can obtain the remaining factor by synthetic division. You should come out with:

$\displaystyle 2a^3+a-18=0~\implies~(a-2)\underbrace{(2a^2+4a+9)}_{\text{qudratic polynomial}}=0$

A product is zero if one factor is zero. The first factor is zero if a = 2. You now have to check if there are any values of a which will make the 2nd factor to be zero. I'll leave that for you.
• Apr 23rd 2014, 08:56 PM
catenary
Re: tricky calculus problem
I see now, the book is actually trying to explain how to solve for the roots of a cubic equation. I was wondering why it had to say "integral root" rather than just "root" because that just confused me. Also when it said "find the latter" i thought it meant 'find the other roots' but actually it meant "the quadratic polynomial". it makes sense that the roots of the quadratic portion should be imaginary since there is only one possible point that fits the picture..
the hardest part was knowing that you must equate the f'(a) to (y2-y1)/(x2-x1) formula for m, which is the hint i think the book should have given..