Calculus I: Tangent line at certain point, second derivative of the curve given. How to solve? 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.
 A: 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,
$$
y = - \frac{1}{20} x^5 - \frac{7}{4} x + \frac{14}{5} .
$$
A: 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!!!
