Question

Help !

Answer 1

and you're asking for help...where...?

Answer 2

you need help with which Question ?

Answer 3

Q2, Q4, Q9 and Q10

Answer 4

Do you still need help with these?

Answer 5

oh yes still cant figure them out

Answer 6

Well, I have some ideas for Q9 and Q10, hope this help
Q9:
\[\int\limits_{0}^{\pi/2}\frac{ d(2\sin \theta) }{ 4(\sin \theta) ^{2} 2\cos \theta }= \int\limits_{0}^{\pi/2}\frac{ 2\cos \theta d \theta }{ 4(\sin \theta) ^{2} 2\cos \theta }= \int\limits_{0}^{\pi/2}\frac{ d \theta }{ 4(\sin \theta)^{2} } \]
\[\theta \rightarrow 0, \sin \theta \rightarrow +\infty\] => Result is infinity
Q10:
The meaning of the integral is the area bounded by x-axis, x = 0, x= 1 and y = 1/sqrt(9 - 4x^2) (actually i dont think this is the answer)
Let u = 2t/3, then
\[\int\limits_{0}^{2/3}\frac{ \frac{ 3 }{ 2 }d(u) }{3\sqrt{1-u ^{2}} }=\frac{ 1 }{ 2 } \int\limits_{0}^{2/3}\frac{ du }{ \sqrt{1-u ^{2}} }\]
Let u = sinθ, 0 <= θ <= arcsin(2/3)
\[=\frac{ 1 }{ 2 }\int\limits_{0}^{\arcsin(\frac{ 2 }{ 3 })}\frac{ d(\sin \theta) }{\cos \theta }\]

Answer 7

sr, at Q9, θ→0,1/sinθ→+∞

Answer 8

This is my solution to Q2:
Determine the equation of the tangent line at x=1 for the ellipse x^2/4+y^2=1:
(from the plot I attached here u can see there are two points A, B such that x=1)
First we have to find the two points in the coordinates at x = 1:
substitute x = 1 in x^2/4+y^2=1
=> 1/4+y^2=1
=> y^2 = 3/4
\[ => y=\pm \sqrt{3}/2\]
\[A=(1,\sqrt{3}/2) , B =(1,-\sqrt{3}/2) \]
Now take the differential of the ellipse function:
(x/2)dx+ 2ydy = 0.
Now, substitute the above coordinates of the points A and B to find the increments.
Thus find the tangent lines at A and B.

Answer 9

For Q4, the slope of tangent line to graph of y = f(x) at any point (x,y) is f'(x). I think this must be mentioned in textbook? Hope you solved it.