core 4 queries?

7) A curve has parametric equations

x = e^2t, y =2t/(1 + t)

(i) Find the gradient of the curve at the point where t = 0
(ii) Find y in terms of x.

9) The points A, B and C have coordinates (1, 3, ?2), (?1, 2, ?3) and (0, ?8, 1) respectively.
(i) Find the vectors AB and AC.
(ii) Show that the vector 2i ? j ? 3k is perpendicular to the plane ABC. Hence find the equation of
the plane ABC.

I tried to do nine by doing the dot product, but i couldnt get it to work?

Anyone give me a clue please? :colondollar:

Reply 1

EEngWillow

Original post by fujitsum

What about question 7, you stuck on both bits?

Yep!

for question 7, show your working

Reply 4

EEngWillow

Original post by fujitsum

Yep!

7)

i) The gradient of a parametric curve would be dy/dt * dt/dx. (Can you find dt/dx?)

ii) You need to find a way of writing this as a cartesian equation, so you'll need to get rid of the t.

9) I assume the question marks are minus signs.

Reply 5

Slowbro93

What are the ? meant to be?

Reply 6

EEngWillow

Original post by fujitsum

9) The points A, B and C have coordinates (1, 3, ?2), (?1, 2, ?3) and (0, ?8, 1) respectively.
(i) Find the vectors AB and AC.
(ii) Show that the vector 2i ? j ? 3k is perpendicular to the plane ABC. Hence find the equation of
the plane ABC.

I tried to do nine by doing the dot product, but i couldnt get it to work?

Anyone give me a clue please? :colondollar:

Assuming you're asking for a clue with part (ii), you're using a.b = |a||b|cos(x), right?

If the plane and the vector are perpendicular, what's the angle between them? How does that affect the RHS?

Reply 7

fujitsum

Yeah, the question marks are minus signs, don't know why its done that.

I am really clueless?? I don't get how to differentiate anything on q7.

Reply 8

EEngWillow

Original post by fujitsum

Yeah, the question marks are minus signs, don't know why its done that.

I am really clueless?? I don't get how to differentiate anything on q7.

\frac{dy}{dx} = \frac{dy}{dt} * \frac{dt}{dx} = \frac{dy}{dt} * \frac{1}{\frac{dx}{dt}}

You'll want to find the derivitive of y with respect to t, and the derivitive of t with respect to x. (This is the same as the reciprocal of the derivitive of x with respect to t).

Do you know how to differentiate e^x and quotients?

(edited 13 years ago)

Reply 9

fujitsum

Original post by EEngWillow

If the plane and the vector are perpendicular, what's the angle between them? How does that affect the RHS?

I dooon't know?? :confused:

Reply 10

fujitsum

Original post by EEngWillow

\frac{dy}{dx} = \frac{dy/dt} * \frac{dt}{dx} = \frac{dy}{dt} * \frac{1}{\frac{dx}{dt}}

Nope, well, i might recognise it, but i don't have any of the assumed knowledge from core 3! :frown:

Reply 11

EEngWillow

Original post by fujitsum

Nope, well, i might recognise it, but i don't have any of the assumed knowledge from core 3! :frown:

Just realised I ****ed up my LaTeX completely, equation is fixed now.

When you differentiate e^5x, you'll get 5e^5x. You have e^2t, so you'll get?

The quotient rule is dy/dx = (vdu - udv)/v^2, where u is the numerator and v the denominator. You use it when you have a fraction where both the numerator and denominator are functions of the variable you're differentiating with respect to.

And for question 9: When two lines are perpendicular, the angle between them is 90 degrees. cos(90) = 0, so if they're perpendicular your scalar product will equal 0.

Reply 12

fujitsum

2e^2t then!

I'll revise quotient rule and have a crack at it then!

How do you do the dot product correctly?

I x'd the two vectors (A&B) together? That doesnt = 0 though?