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I am studying Strang's Calculus text and find myself stumped by the following problem:

'Find a curve that is tangent to y = 2x - 3 at x = 5. Find the normal line to that curve at (5, 7).'

All my searching only yields results for finding tangent lines to a curve, not the other way around.

I thought to take the antiderivative of the function and substitute the point of tangency back into the equation to solve for the resulting constant, but this didn't give me the correct answer.

Suggestions on how to proceed?

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  • $\begingroup$ What answer did you get - there are many! $\endgroup$
    – tomi
    Commented Nov 2, 2015 at 15:56
  • $\begingroup$ For the latter question, the normal line has slope equal to $-1/2$. Then substitute $(5,7)$ into the normal line. $\endgroup$
    – James Pak
    Commented Nov 2, 2015 at 15:56
  • $\begingroup$ Tangency (in this context) only requires that the figures involved "touch" each other at exactly one point, at least locally. Tangent lines are useful in visualizing derivatives and points of tangency, but tangency is not limited to straight lines only. $\endgroup$
    – Corellian
    Commented Nov 2, 2015 at 16:20
  • $\begingroup$ @tomi: I got an equation: $y = x^2-4x-3$; when I graphed this on the same plot as $y=2x-3$, the two lines intersected but were not tangent to one another. $\endgroup$
    – anatta
    Commented Nov 2, 2015 at 18:22

3 Answers 3

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They want "a curve" - because there are an infinite number of curves that will satisfy the conditions!

So you need $y=7$ when $x=5$ and $\frac {dy}{dx}=2$ when $x=5$.

The simplest kind of curve would be a quadratic; I would suggest $y=x^2+ax+b$.

Substitute the known conditions into that curve to create simultaneous equations for $a$ and $b$.

Solve and there you are!

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  • $\begingroup$ So if I substitute (5,7) into the quadratic equation, I get 5a + b = -18 ; how to I find another simultaneous equation? $\endgroup$
    – anatta
    Commented Nov 2, 2015 at 18:18
  • $\begingroup$ Differentiating the expression for $y$ gives $\frac {dy}{dx}=2x+a$. You can substitute $\frac {dy}{dx}=2$ and $x=5$ to get $2=10+a$ $\endgroup$
    – tomi
    Commented Nov 3, 2015 at 14:43
  • $\begingroup$ Ah! Thank you for the explanation. $\endgroup$
    – anatta
    Commented Nov 3, 2015 at 18:38
  • $\begingroup$ The solution given in the back of the book for a tangent curve is: $y=(1/5)x^2+2$. Do you know how one might easily arrive at this equation? $\endgroup$
    – anatta
    Commented Nov 3, 2015 at 18:40
  • $\begingroup$ Yes. They have started with $y=ax^2+b$, which gives $\frac{dy}{dx}=2ax$. Substituting in the same values as I used earlier for $x$, $y$ and $\frac {dy}{dx}$ gives you $7=25a+b$ and $2=10a$. You shoudl be able to do it from there. $\endgroup$
    – tomi
    Commented Nov 3, 2015 at 20:56
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Choose any function $f(x)$ such that sensible values for $f(5)$ and $f'(5)$ exist.

We want to create a new function $g(x)$ such that $g(5)=7$ and $g'(5)=2$.

Let $g(x)=af(x)+b \Rightarrow 7=af(5)+b$

Differentiation gives $g'(x)=af'(x) \Rightarrow 2=af'(5)$

So $a=\frac 2 {f'(5)}$

And $7=\frac 2 {f'(5)}f(5)+b$

So $b=7-\frac {2f(5)} {f'(5)}$

Which gives $g(x)=\frac {2f(x)} {f'(5)}+7-\frac {2f(5)} {f'(5)}$

Or $g(x)=\frac {2f(x)} {f'(5)}-\frac {2f(5)} {f'(5)}+7$

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HINT

Take any curve involving two constants.

Find two intersection points with the given line

$$ \frac{x}{3/2}+\frac{y}{3}+=1 $$

by solving the two together.

Relate the two constants so the points coincide to a single tangent point, involving only a single constant.

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