# C4 Mathematics - parametric equations question.Watch

#1
Question 19)c) of Mixed exercise 4F of the Edexcel C4 textbook is the one I'm after.

Here is the full question (I've done the first two parts):

19)

a) Given that x = 2^t, by using logarithms prove that

dx/dt = 2^t.ln2

A curve C has parametric equations x = 2^t and y = 3t^2. The tangent to C at the point with coordinates (2,3) cuts the x-axis at the point P.

b) Find dy/dx in terms of t.

c) Calculate the x-coordinate of P, giving your answer to 3 decimal places.

I know it is probably really easy to solve but my mind has gone blank. All I know is that if it crosses the x axis then the y coordinate is 0, and I need to figure out what t is somehow.
0
7 years ago
#2
Using the x cordinate of the tangent, you can find the value of t. Sub that into dy/dx to find the gradient of the tangent. Find the equation passing through points (2,3) and after let y=0 to find the x-cordinate of the point P.
0
7 years ago
#3
(Original post by A level Az)
Question 19)c) of Mixed exercise 4F of the Edexcel C4 textbook is the one I'm after.

Here is the full question (I've done the first two parts):

19)

a) Given that x = 2^t, by using logarithms prove that

dx/dt = 2^t.ln2

A curve C has parametric equations x = 2^t and y = 3t^2. The tangent to C at the point with coordinates (2,3) cuts the x-axis at the point P.

b) Find dy/dx in terms of t.

c) Calculate the x-coordinate of P, giving your answer to 3 decimal places.

I know it is probably really easy to solve but my mind has gone blank. All I know is that if it crosses the x axis then the y coordinate is 0, and I need to figure out what t is somehow.
Find the value of t that satisfies and simultaneously (i.e. find the value of t that represents the parameter of the point (2,3)) and then plug that into your gradient function from part (b) to get the gradient of the tangent to the curve at (2,3). Use the information to find the equation of that tangent and find the value of x for which y=0 for the tangent, not the curve.
0
#4
Thanks a bunch guys
0
4 years ago
#5
I'm stuck on this question because i'm trying to get the equation into the form y=mx+c to find c then find the x coordinate when y=0 and I keep getting the wrong answer.
0
4 years ago
#6
(Original post by jit987)
Using the x cordinate of the tangent, you can find the value of t. Sub that into dy/dx to find the gradient of the tangent. Find the equation passing through points (2,3) and after let y=0 to find the x-cordinate of the point P.
how do you get the equation of the line using y=mx+c I keep getting it wrong
0
4 years ago
#7
(Original post by Milliesgreen)
I'm stuck on this question because i'm trying to get the equation into the form y=mx+c to find c then find the x coordinate when y=0 and I keep getting the wrong answer.
show what you have as your y=mx+c
0
4 years ago
#8
(Original post by TenOfThem)
show what you have as your y=mx+c
I got y=(6/2ln2)x +c
so x=2 and y=3
3=2(6/2ln2) + c
then whatever i do after that gets me the wrong answer
0
4 years ago
#9
(Original post by TenOfThem)
show what you have as your y=mx+c
don't worry i've got it!
0
4 years ago
#10
(Original post by Milliesgreen)
I got y=(6/2ln2)x +c
so x=2 and y=3
3=2(6/2ln2) + c
then whatever i do after that gets me the wrong answer
So you got

x = 2 - ln2

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