1. Differentiation Help!

Can someone help me on the following questions cos im just so stuck!

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

2. 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?

3. 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) $\displaystyle \nabla{f(x)}=\frac{\partial{f}}{\partial{x}}\hat{x }$

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?

4. Gradient is a synonym for "slope of the tangent at the
point in question" in this context.

RonL

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

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
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)

6. 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.

7. "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.

8. 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.