Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have never seen this problem in my homework.. I need help right before final!!

Let $\frac{d^2y}{dx^2}=y''=-x^3$ at every point on a curve. The equation of the tangent line at $(1,1)$ is $y = 3-2x$. Find the equation of the curve.

If someone can help me out with this problem, will greatly appreciate. Thank you.

share|improve this question
    
Hope my answer clears things up! Good luck preparing for your final. –  izœc Mar 24 at 5:00

2 Answers 2

up vote 1 down vote accepted

Observe that via integration, it follows that $$ y' = - \frac{x^4}{4} + C. $$ Hence, at $(1,1)$, $y'(1) = C - \frac{1}{4}$, and the tangent line is of the form $y - 1 = ( C - \frac{1}{4} )(x - 1)$, that is ( using the knowledge of the tangent curve at (1,1) ) that $$ 3 - 2x = y = \left(C- \frac{1}{4} \right)x + \frac{5}{4} - C. $$ Then, $C - \frac{1}{4} = - 2 \implies C = - \frac{7}{4}$. Also, we know that $\frac{5}{4} - C = 3 \implies C = - \frac{7}{4}$. Since these conditions agree, we have the solutions that $$ y' = - \frac{1}{4} \left(x^4 + 7 \right) $$ so that $y$ is any curve of the form $y = - \frac{1}{20} x^5 - \frac{7}{4} x + K$, for real numbers $K$. Further, we know that $(1,1)$ is a point on the curve, and hence that $K$ must be such that $1 = - \frac{1}{20} - \frac{7}{4} + K$. It follows that $K = \frac{56}{20} = \frac{14}{5}$. So, $$ y = - \frac{1}{20} x^5 - \frac{7}{4} x + \frac{14}{5} . $$

share|improve this answer
1  
Thank you for the step-by-step solution. Now I get it. –  Kibbles Mar 24 at 5:07
    
@Kibbles No problem, I am glad it helped. –  izœc Mar 24 at 5:38

Since

$y'' = -x^3, \tag{1}$

the general form of $y'$ is, by integration,

$y' = -\dfrac{1}{4}x^4 + a, \tag{2}$

and so the general form of $y(x)$ must be

$y = -\dfrac{1}{20}x^5 + ax + b; \tag{3}$

since the curve (3) passes through the point $(1, 1)$ we have

$1 = -\dfrac{1}{20} + a + b; \tag{4}$

an since the slope at this point is $y'(1) = -2$ we also have

$-2 = -\dfrac{1}{4} + a \tag{5}$

or

$a = -\dfrac{7}{4}. \tag{6}$

When (6) is used in (4) we find

$b = 1 + \dfrac{1}{20} + \dfrac{7}{4} = \dfrac{14}{5}; \tag{7}$

thus we have

$y(x) = - \dfrac{1}{20}x^5 - \dfrac{7}{4}x + \dfrac{14}{5}. \tag{8}$

A race to the finish with izoec! Some problems only have one answer!

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.