Author Topic: Differentiation (Read 6607 times) Tweet Share

d0minicz · « **Reply #15 on:** May 13, 2009, 02:54:25 pm »

For the curve with the equation $y= 2x^3 -9x^2 +12x +8$ find the values of x for which:
a) $\frac {dy}{dx} < 0$ and $\frac {d^2y}{dx^2} > 0$
b) $\frac {dy}{dx} < 0$ and $\frac {d^2y}{dx^2} < 0$
thanks..

Damo17 · « **Reply #16 on:** May 13, 2009, 03:18:12 pm »

Quote from: d0minicz on May 13, 2009, 02:54:25 pm

For the curve with the equation $y= 2x^3 -9x^2 +12x +8$ find the values of x for which:
a) $\frac {dy}{dx} < 0$ and $\frac {d^2y}{dx^2} > 0$
b) $\frac {dy}{dx} < 0$ and $\frac {d^2y}{dx^2} < 0$
thanks..

a)
$\frac{dy}{dx}=6x^2-18x+12$

$\frac{d^2y}{dx^2}=12x-18$

to find the values of x that satisfy $\frac {dy}{dx} < 0$ and $\frac {d^2y}{dx^2} > 0$ solve the inequality $\frac {d^2y}{dx^2} > 0$ .

$\therefore 12x-18>0$
$12x>18$
$x>\frac{3}{2}$

b) do the same but solve the inequality $\frac {d^2y}{dx^2} < 0$

$\therefore 12x-18<0$
$x<\frac{3}{2}$

EDIT: For part a you would have to exclude $2$ as it would make $\frac{dy}{dx}=0$ . For part b you would exclude $1$ and that would also make $\frac{dy}{dx}=0$ . These values of 1 and 2 are worked out when you solve $\frac{dy}{dx}=0$ .

$\frac{dy}{dx}=6(x-2)(x-1)$

d0minicz · « **Reply #17 on:** May 15, 2009, 08:57:26 pm »

A cylindrical tank 5m high is base radius 2m is initially full of water. Water flows through a hole at the bottom of the tank at the rate of $\sqrt h$ $m^3/hour$ , where h metres is the depth of the water remaining in the tank after t hours. Find:
a) $\frac {dh}{dt}$
b)i) $\frac {dV}{dt}$ when $V= 10\pi m^3$
ii) $\frac {dh}{dt}$ when $V= 10\pi m^3$
thanks

Damo17 · « **Reply #18 on:** May 15, 2009, 09:17:25 pm »

Quote from: d0minicz on May 15, 2009, 08:57:26 pm

A cylindrical tank 5m high is base radius 2m is initially full of water. Water flows through a hole at the bottom of the tank at the rate of $\sqrt h$ $m^3/hour$ , where h metres is the depth of the water remaining in the tank after t hours. Find:
a) $\frac {dh}{dt}$
b)i) $\frac {dV}{dt}$ when $V= 10\pi m^3$
ii) $\frac {dh}{dt}$ when $V= 10\pi m^3$
thanks

a) $\frac{dV}{dt}=-\sqrt{h}$

$V=\pi r^2h$

as $r=2$
$V=4\pi h$

$\frac{dV}{dh}=4\pi$ $\frac{dh}{dV}=\frac{1}{4\pi}$

$\frac{dh}{dt}=\frac{dV}{dt} \cdot \frac{dh}{dV}$
$=-\sqrt{h} \cdot \frac{1}{4\pi}$
$= -\frac{\sqrt{h}}{4\pi}$

b)i) $\frac{dV}{dt}=-\sqrt{h}$

as $V=10\pi$

$10\pi=4\pi h$

$h=\frac{10}{4}$

$\therefore \frac{dV}{dt}=-\sqrt{\frac{10}{4}}$
$= -\frac{\sqrt{10}}{2}$ $m^3/h$

ii) $\frac{dh}{dt}= -\frac{\sqrt{\frac{10}{4}}}{4\pi}$
$= -\frac{\sqrt{10}}{8\pi}$ m/h

d0minicz · « **Reply #19 on:** May 24, 2009, 02:04:23 pm »

The equation of a curve is $2x^2 + 8xy + 5y^2 = -3$ . Find the equation of the two tangents which are parallel to the x-axis.
thx =]

kamil9876 · « **Reply #20 on:** May 24, 2009, 02:48:54 pm »

Quote from: d0minicz on May 24, 2009, 02:04:23 pm

The equation of a curve is $2x^2 + 8xy + 5y^2 = -3$ . Find the equation of the two tangents which are parallel to the x-axis.
thx =]

differentiating all terms:
$<br />4x+8\frac{d}{dx}(xy)+\frac{d}{dx}(5y^2)=0$
first differentiatiation is requires product rule, second is a chain rule:

$4x+8(\frac{d}{dx}(x)y+\frac{d}{dx}(y)x)+\frac{dy}{dx}\frac{d(5y^2)}{dy}=0$
$4x+8(y+\frac{dy}{dx}x)+10y\frac{dy}{dx}=0$

now parralel to x axis means $\frac{dy}{dx}=0$ hence sub that in to get a second equation. Now you have two simultaneous equations to solve. Solve for y because equations that are parralel to the x axis are of the form y=c where c is a constant:

4x+8y=0
x=-2y

sub that back into the original as it gives you a quadratic.

d0minicz · « **Reply #21 on:** June 16, 2009, 05:32:52 pm »

Find the coordinates of the points of inflexion of y= sinx for x=[0,2Pi]
thanks.

TrueTears · « **Reply #22 on:** June 16, 2009, 05:37:56 pm »

I'm sure you can do that question. I think you are confused because book's answer is wrong.

I clearly remember this question.

d0minicz · « **Reply #23 on:** June 16, 2009, 05:39:48 pm »

wats the answer sposed to be ?

d0minicz · « **Reply #24 on:** June 16, 2009, 05:45:16 pm »

Let $g(x) = e^{-x} f(x)$ where f is a differentiable function for all the real numbers. There is a point of inflexion on the graph of $y= g(x)$ at (a,g(a)). An expression for f''(a) in terms of f'(a) and f(a) is:
thanks.

TrueTears · « **Reply #25 on:** June 16, 2009, 05:52:17 pm »

Quote from: d0minicz on June 16, 2009, 05:39:48 pm

wats the answer sposed to be ?

Find second derivative, set it to 0, solve for the required domain, and you are done. (You should check that the sign changes on both side by picking values and subbing it into second derivative, but here it is obviously they change)

d0minicz · « **Reply #26 on:** June 16, 2009, 06:34:39 pm »

Find the derivative of : $y= tan^{-1} (\frac {1}{3x})$
thank you

dcc · « **Reply #27 on:** June 16, 2009, 06:38:42 pm »

Quote from: d0minicz on June 16, 2009, 05:45:16 pm

Let $g(x) = e^{-x} f(x)$ where f is a differentiable function for all the real numbers. There is a point of inflexion on the graph of $y= g(x)$ at (a,g(a)). An expression for f''(a) in terms of f'(a) and f(a) is:
thanks.

First, note that $g(a) = e^{-a}f(a)$

Using the product rule, we find that: $g'(x) = e^{-x} f'(x) - e^{-x} f(x)$

Again, we find $g''(x) = e^{-x} f''(x) - 2e^{-x} f'(x) + e^{-x} f(x)$

$g''(a) = 0 = e^{-a} f''(a) - 2e^{-a}f'(a) + e^{-a}f(a) \implies f''(a) = 2f'(a) - f(a)$

dcc · « **Reply #28 on:** June 16, 2009, 06:44:21 pm »

Quote from: d0minicz on June 16, 2009, 06:34:39 pm

Find the derivative of : $y= \arctan \left( \frac{1}{3x} \right)$
thank you

Let $u = \frac{1}{3x} \implies \frac{du}{dx} = -\frac{1}{3x^2}$

$y = \arctan \left( u \right) \implies \frac{dy}{du} = \frac{1}{1 + u^2}$

Therefore by the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \dfrac{du}{dx} = \frac{1}{1 + \left( \frac{1}{3x} \right)^2} \cdot \left(-\frac{1}{3x^2} \right) = - \frac{3}{1 + 9x^2}$

Over an appropriate domain, of course.

d0minicz · « **Reply #29 on:** June 16, 2009, 07:57:11 pm »

$\int x \sqrt {4-x} dx$
need to check if the answer is right ta =]

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