# Differentiation Help!

• Nov 11th 2005, 08:14 AM
Confuzzled?
Differentiation Help!
Can someone help me on the following questions cos im just so stuck! :confused:

1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2

2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b.

3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0
Find the rate at which the radius is changing when t=1 and when t=2.

Any help will be much appreciated! Thanks in advance :cool:
• Nov 11th 2005, 08:37 AM
Jameson
Quote:

Originally Posted by Confuzzled?
1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2

I'll start with the first one. In order to find the tangent line and the normal line, you need a point and a slope. The point you can find by plugging in your x value, and how would you find the slope? How can you rewrite xy=4 into something easier to differentiate?
• Nov 11th 2005, 08:44 AM
Jameson
Quote:

Originally Posted by Confuzzled?
2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b.

3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0

For the second one, I don't know why a function of one variable makes a reference to gradient, as one normally sees that with functions of three variables. But, I guess you could define the gradient for f(x) as:

(1) $\nabla{f(x)}=\frac{\partial{f}}{\partial{x}}\hat{x }$

and work from there. It's really just the derivative for this function.

And for the third one, I don't even see a question being asked. Inforrmation is provided, what are they looking for?
• Nov 11th 2005, 10:50 AM
CaptainBlack
Gradient is a synonym for "slope of the tangent at the
point in question" in this context.

RonL
• Nov 11th 2005, 11:25 AM
Confuzzled?
Quote:

Originally Posted by Confuzzled?
Can someone help me on the following questions cos im just so stuck! :confused:

1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2

2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b.

3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0

Any help will be much appreciated! Thanks in advance :cool:

sorry guys i was kinda in a rush. Anyways the actual question for no. 3 was

3. Find the rate at which the radius is changing when t=1 and when t=2.

And Jameson, I converted xy=4 into y=4x^-1 (is that right?) I worked with that yet I didnt get the right answer in the end (the answers are at the back)
• Nov 11th 2005, 11:27 AM
Confuzzled?
oh and for question 2 I think have been able to attempt a similar question where I had to form a simultaneous equation to find a and b but I wasnt sure how I could do this for that question.
• Nov 11th 2005, 11:33 AM
ticbol
"1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2 "

Slope of tangent line is dy/dx.
Slope of normal line is the negative reciprocal of dy/dx.

xy = 4
Differentiate both sides with respect to x,
x[dy/dx] +y[dx/dx] = 0
dy/dx = -y/x

When x=2:
xy = 4
2y = 4
y = 2 also.
So, dy/dx = -y/x = -2/2 = -1 -----slope of tangent line.
And so, slope of normal line = -1/(-1) = 1 ----------------***

Then, for equation of tangent line at point (2,2), using the point-slope form,
(y-y1) = m(x-x1)
Or, in its slope-intercept form or y=mx+b,
y -2 = -x +2

The normal line at (2,2):
y-2 = 1(x-2) -----point-slope form, answer.
Or, y = x -------slope-intercept form, answer.

-----------------------
" 2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b. "

y = ax^2 +bx -a^2 -------(i)
dy/dx = 2ax +b = -4
At point (-2,-13),
2a(-2) +b = -4
-4a +b = -4
b = 4a -4 ------(1)

Two unknowns, a&b, we need another equation.
y = ax^2 +bx -a^2 -------(i)
At point (-2,-13),
-13 = a(-2)^2 +b(-2) -a^2
-13 = 4a -2b -a^2 ----------(2)
Substitute into (2) the b from (1)
-13 = 4a -2(4a -4) -a^2
-13 = 4a -8a +8 -a^2
a^2 +4a -21 = 0
(a+7)(a-3) = 0
a = -7 or 3 -------------------answer.

When a= -7,
b = 4a -4 = 4(-7) -4 = -32
When a = 3,
b = 4(3) -4 = 8
Hence, b = -32 or 8 -----------------answer.

------------------
" 3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0 "

If I can understand that, I can answer it.
• Nov 11th 2005, 11:57 AM
ticbol
Ah, so there is the correction for #3.

I am using a mac Powerbook now because my Dell laptop was damaged by ligtning storm a few days ago. I am new at Mac OS so I am really very slow in answering now. (I don't even know how to use the Clipboard!).
The "Quote" feature above doesn't seem to work with Mac.

" 3. Find the rate at which the radius is changing when t=1 and when t=2 "
" 3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0 "

r = (t^3)/3 -2t
Differentiate both sides with respect to time t,
dr/dt = t^2 -2 ------rate of change of the radius.
So,
When t=1, dr/dt = (1)^2 -2 = -1 -------answer.
When t=2, dr/dt = (2)^2 -2 = 2 -------answer.