Help with calc questions. (1 Viewer)

Nake · Jun 22, 2009

just been doing a few questions, wondering if someone could post up some help...

1)

A curve has y' = 6x^2 + 12x - 5 . If the curve passes through the point (2,-3), find the equation of the curve.

2)

The height in meters of an object thrown up into the air is given by h = 20t - 2t^2 where t is time in seconds. Find the maximum height that the object reaches.

I diff'd to find a=4 , but stuffed if i know where to go from there. Cause if you make t = 0, h will also = 0 ...

3)

For the curve y = 3x^4 + 8x^3 - 48x^2 - 1
a) find any stationary points and determine their nature
b) Sketch the curve

diff'd it, didnt know how to find out what x = with 3 terms being there

Any help would be loverly

Drongoski · Jun 22, 2009

I'll do Q1 & 2

$\80dpi Q1. \\ \\ \ \ {y}'\ =\ 6x^{2}+12x-5\ \ \Rightarrow \ \ y\ =\ \int (6x^{2}+12x-5)\ dx\\ \\ \ =\ \ 2x^{3}+6x^{2}-5x+C\\ \\ This\ curve\ passes\ thru\ (2,-3)\ means\ when\ x=2,\ y=-3\\ \\ \therefore \ \ 2\times 2^{3}\ +\ 6\times 2^{2}\ -\ 5\times 2\ +\ C\ =\ -3\ \ \Rightarrow \ \ C\ =\ -33\\ \\ \\ Q2.\\ \\ h\ =\ 20t-2t^{2}\ =\ 2t(10-t)\ \Rightarrow \ h=0\ at\ t=0\ and\ t=10\\ \\ Max \ h\ occurs\ at\ mid-point\ i.e.\ at\ t=5\ \ (since\ h\ is\ a\ quadratic\ in\ t)\\ \\ h_{max}\ \ = \ 2\times 5(10-5)\ =\ 50\ \ units$

Drongoski · Jun 22, 2009

Q2 by calculus

$\80dpi h(t)\ =\ 20t-2t^{2}\ \ \Rightarrow \ \ \frac{\mathrm{d} h}{\mathrm{d} t}\ =\ 20-4t\ =\ 0\ \ \ @\ t=5\\ \\ \frac{\mathrm{d^{2}}h }{\mathrm{d} t^{2}}\ =\ -4\ \ \ (negative\ whatever\ value\ of\ t)\ \Rightarrow \ we\ have\ a \ max\ h\ @\ \ t=5\\ \\ h_{max}\ =\ 20\times 5\ -\ 2\times 5^{2}\ =\ 50\ m$

Aquawhite · Jun 22, 2009

Q3.

$y = 3x^4 + 8x^3 - 48x^2 - 1$

Stat points occur when dy'dx = 0

$y' = 12x^{3}+24x^{2}-96x$ , thus:

$y' = 12x^{3}+24x^{2}-96x = 0$

where 12x can be taken out as commong factor.

$12x(x^{2}+2x-8) = 0 \Rightarrow 12x(x-4)(x+2)$ because we have a quadratic.

therefore there are stationary points where x = 0, x = 4 and x = -2
sub these back into the initial equation for the y-coordinates (I'm too lazy to that right now cause I'm missing GNW lol)

For their nature, use second derivative using the values of the stationary points of x.

y'' using x=0, x=4 and x=-2

$y''=36x^{2}+48x-96$ where y'' < 0 (maximum), where y'' > 0 (minimum)

thus x=0 is a maximum
x=4 is minimum
x=-2 is minimum

Sorry for not finding y coordinates, but too lazy. Also if it's wrong... I apologise.

P.S. I have no idea why I used LaTeX for this but I guess I better get used to it.

kurt.physics · Jun 22, 2009

Nake said:
3)

For the curve y = 3x^4 + 8x^3 - 48x^2 - 1
a) find any stationary points and determine their nature
b) Sketch the curve

diff'd it, didnt know how to find out what x = with 3 terms being there

Any help would be loverly

$\frac{dy}{dx} = 12x^3 + 24x^2 - 96x$

For stationary points, $\frac{dy}{dx} = 0$

ie:

$12x^3 + 24x^2 - 96x = 0$

$12x\left(x^2 + 2x - 8\right) = 0$

$12x = 0$

$\therefore x = 0$

or

$x^2 + 2x - 8 = 0$

$\left(x +4\right)\left(x-2\right) = 0$

$\therefore x = -4, 2$

Thus we have the solutions

$x = 0, 2, -4$

$\left(0,-1\right)$ is a stationary point

$\left(2, -81\right)$ is a stationary point

$\left(-4, -513\right)$ is a stationary point

$\frac{d^2y}{dx^2} = 36x^2 + 48x - 96$

when $x = 0$ , $\frac{d^2y}{dx^2} = -96$

$\therefore \frac{d^2y}{dx^2} < 0$

$\therefore \left(0, -1\right)$ is a maxium point

when $x=2$ , $\frac{d^2y}{dx^2} = 144$

$\therefore \frac{d^2y}{dx^2} > 0$

$\therefore \left(2, -81\right)$ is a minimum point

when $x=-4$ , $\frac{d^2y}{dx^2} = 288$

$\therefore \frac{d^2y}{dx^2} > 0$

$\therefore \left(-4, -513\right)$ is a minimum point

For sketching the curve:

y-intercept occurs when x = 0

plot the minimum and max points

Aquawhite · Jun 22, 2009

damn, you went to the effort for the y-coordinates! *pirate noise*

I couldn't be bothered hunting for the calculator

kurt.physics · Jun 22, 2009

Aquawhite said:
Q3.

$y = 3x^4 + 8x^3 - 48x^2 - 1$

Stat points occur when dy'dx = 0

$y' = 12x^{3}+24x^{2}-96x$ , thus:

$y' = 12x^{3}+24x^{2}-96x = 0$

where 12x can be taken out as commong factor.

$12x(x^{2}+2x-8) = 0 \Rightarrow 12x(x-4)(x+2)$ because we have a quadratic.

therefore there are stationary points where x = 0, x = 4 and x = -2
sub these back into the initial equation for the y-coordinates (I'm too lazy to that right now cause I'm missing GNW lol)

For their nature, use second derivative using the values of the stationary points of x.

y'' using x=0, x=4 and x=-2

$y''=36x^{2}+48x-96$ where y'' < 0 (maximum), where y'' > 0 (minimum)

thus x=0 is a maximum
x=4 is minimum
x=-2 is minimum

Sorry for not finding y coordinates, but too lazy. Also if it's wrong... I apologise.

P.S. I have no idea why I used LaTeX for this but I guess I better get used to it.

Beat me to it

kurt.physics · Jun 22, 2009

Aquawhite said:
damn, you went to the effort for the y-coordinates! *pirate noise*

I couldn't be bothered hunting for the calculator

lol

Help with calc questions. (1 Viewer)

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