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# p3 parametrics watch

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1. hey, stuck on a question :

x = a sec t
y = a tan t, and i need to find dy/dx

sounds easy, but i dont know where to start from, to differentiate x and y, then apply the chain rule.

anyone help plz?
2. (Original post by Taslima)
hey, stuck on a question :

x = a sec t
y = a tan t, and i need to find dy/dx

sounds easy, but i dont know where to start from, to differentiate x and y, then apply the chain rule.

anyone help plz?
cosect.
3. (Original post by Ralfskini)
cosect.
welldone, how did yu get it?
• Find dx/dt (put secx as 1/cosx and use quotient rule)

• Find dy/dt

• Then dy/dx = (dy/dt)/(dx/dt)

The answer you should get is cosec(t)
4. basically have to use the rule:

dy/dx = (dy/dt)/(dx/dt)

y=atant
dy/dt = a (sect)^2

x= asect
dx/dt = asect tant

dy/dx = (asec^2t)/(asecttant)
= sect/tant
=1/(cost.tant)
=1/sint
=cosect

Hope that helps
5. (Original post by el GaZZa)
• Find dx/dt (put secx as 1/cosx and use quotient rule)

• Find dy/dt

• Then dy/dx = (dy/dt)/(dx/dt)

The answer you should get is cosec(t)
this doesnt help me at all.. what happens to the a's? in both x and y, and how do i "find" y = a tan t, its the whole reason why im posting, so i can be told "how" to do it..
6. (Original post by Taslima)
this doesnt help me at all.. what happens to the a's? in both x and y, and how do i "find" y = a tan t, its the whole reason why im posting, so i can be told "how" to do it..
a's cancel out when you divide one by the other
7. (Original post by Silly Sally)
basically have to use the rule:

dy/dx = (dy/dt)/(dx/dt)

y=atant
dy/dt = a (sect)^2

x= asect
dx/dt = asect tant

dy/dx = (asec^2t)/(asecttant)
= sect/tant
=1/(cost.tant)
=1/sint
=cosect

Hope that helps
wow! thanks very very much!
8. hey, i have another

x = t cos t
y = t sin t

find dy/dx, any helpers plz?
9. I posted it like that as an outline to parametric differentiation (essentially the rule that dy/dx = (dy/dt)/(dx/dt). The rest is standard differentiation (and I thought you were asking for help with the parametric part).

Ok, well firstly, a is a constant, which is multiplied by the rest of the function. Because the function is being differentiated with respect to t, a is just a number and nothing happens to it. For example, d(3x^2)/dx = 3*(2x). This is exactly the same as if we differentiated x^2 and multiplied by 3 afterwards. So if you like, ignore the a while differentiating, and then multiply the answer by a afterwards.

To find d(tanx)/dx, split tanx into sinx/cosx and apply the quotient rule. (Let u=sinx, v=cosx and plug into [v(du/dx) - u(dv/dx)]/(v^2)] ). Then do the same for secx (which is 1/cosx).

It will benefit anyone's calculus skills to be able to do this for all 6 trigonometric functions, I suggest you take a few minutes to do this as it will give you a sounder base for calculus questions in the exam.

Incidentally, the differential of tanx is used often and so its a good idea to learn it without having to work it out, again in case it comes up in exams.

Let me know what answers you get
10. This time you've got to use the product rule [v(du/dx) + u(dv/dx)].

(btw, I'm not trying to skimp on the workload or anything, but it's more beneficial to you if you were to do it!) Let me know what you get.
11. (Original post by el GaZZa)
I posted it like that as an outline to parametric differentiation (essentially the rule that dy/dx = (dy/dt)/(dx/dt). The rest is standard differentiation (and I thought you were asking for help with the parametric part).

Ok, well firstly, a is a constant, which is multiplied by the rest of the function. Because the function is being differentiated with respect to t, a is just a number and nothing happens to it. For example, d(3x^2)/dx = 3*(2x). This is exactly the same as if we differentiated x^2 and multiplied by 3 afterwards. So if you like, ignore the a while differentiating, and then multiply the answer by a afterwards.

To find d(tanx)/dx, split tanx into sinx/cosx and apply the quotient rule. (Let u=sinx, v=cosx and plug into [v(du/dx) - u(dv/dx)]/(v^2)] ). Then do the same for secx (which is 1/cosx).

It will benefit anyone's calculus skills to be able to do this for all 6 trigonometric functions, I suggest you take a few minutes to do this as it will give you a sounder base for calculus questions in the exam.

Incidentally, the differential of tanx is used often and so its a good idea to learn it without having to work it out, again in case it comes up in exams.

Let me know what answers you get
aww! how sweet!! thank00 , that was very nice of yu, thankyu alot! i didnt intend on sounding mean hehe, but i got to sec t/tan t, on my own, until silly sally showed me the rest of it but, thankyu alot! , but ive learnt that if there's a constant infront, it is just normal differentiation and ignore the a, but if there's a variable, then it's product rule
12. Glad I could be of use
13. (Original post by el GaZZa)
Glad I could be of use
weeee!! just done it

i gots me : (tcost + sint)/(cost - tsint)

i hope thats right!
14. (Original post by Taslima)
weeee!! just done it

i gots me : (tcost + sint)/(cost - tsint)

i hope thats right!
15. (Original post by el GaZZa)
w00000!! go tazi!
16. okie, i have another question!! :P

i need to find the equation of the normal to the curve with parametric equations
x = 3t - t^3
y = t^2
at the point where t = 2

this is what i have so far :

dx/dt = 3 - 3t^2
dy/dt = 2t

=> dy/dx = 2t/(3 - 3t^2)
at [ t = 2 ], gradient is -4/9, therefore the normal gradient will be 9/4.

now for the normal equation..

y - y1 = m (x - x1)
y - _ = 9/4 (x - _)

how do i know what the x and y values are to put into the equation?
17. (Original post by Taslima)
okie, i have another question!! :P

i need to find the equation of the normal to the curve with parametric equations
x = 3t - t^3
y = t^2
at the point where t = 2

this is what i have so far :

dx/dt = 3 - 3t^2
dy/dt = 2t

=> dy/dx = 2t/(3 - 3t^2)
at [ t = 2 ], gradient is -4/9, therefore the normal gradient will be 9/4.

now for the normal equation..

y - y1 = m (x - x1)
y - _ = 9/4 (x - _)

how do i know what the x and y values are to put into the equation?
use the t value that you`ve used into the original parametric equation
in your case its t = 2
then
x = 3(2) - (2)^3
y = 2^2

x = -2
y = 4

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