Answer to Question #186973 in Calculus for Hyeri

Question

1. Write the equation of the curve for which y'''=-96x, if the curve is tangent to the line 20x+y+1=0 at (1, -3) and passes thru (0, 1)?

2. A particle moves such that its acceleration is defined by the equation a=3t²-2. When t=0, s=5 and when t=1, v=-20. What is s when t=2?

3. A sandbag is dropped from a balloon rising at 9.6 m/s at a height of 64 m. Find the highest point reached by the sandbag and the time of flight. Note, use g=-10 m/s².

Answer 1

Ans:-$\Rightarrow a=3t²-2$

$\dfrac{dv}{dt}=3t^2-2\\$

integrating with respect to t

$v=t^3-2t+c_1$

at

$t=1, v=-20 \to c_1=-19\\ \Rightarrow v=t^3-2t-19$

$\Rightarrow \dfrac{ds}{dt}=t^3-2t-19$

again integrate with respect to t

$\Rightarrow s=\dfrac{t^4}{4}-t-19t+c_2$

at $t=0, s=5 \to c_2 =5$

then required equation for $s=\dfrac{t^4}{4}-t-19t+5\\$

then at $t=2 ,s=-31$

(3) Ans:-

$\Rightarrow x=ut+\dfrac{1}{2}at^2\\ \Rightarrow -64=9.6+\dfrac{1}{2}\times (-10)\times t^2\\ \Rightarrow 5t^2-9.6-64=0\\ \Rightarrow t=4.66sec$

So, the time of flight of sandbag will be $4.66 sec$

Displacement of balloon after $4.66 sec =9.6*4.66=44.736$ m

Balloon was at 64m height when sandbag was dropped

Highest height delivered by the sandbag will be$64+44.736=108.736m$

(1)Ans:-

$y'''=-96x\\$

integrate with respect to x

$\dfrac{d^2y}{dx^2}=-48x^2+c_1\\$

$\dfrac{dy}{dx}=-16x^3+c_1x+c_2\\$

we know that the curve is tangent to the line $y=-20x-1$ then

$\dfrac{dy}{dx}=-20=-16x^3+c_1x+c_2$

then put$(0,1)$ and $(1,-3)$ then equation will be

$\dfrac{dy}{dx}=-16x^3+16x-20$

again integrate with respect to $x$

$y=4x^4+8x^2-20x$