A curve C is defined by the parametric equations x=(4-e^(2-6t))/4 , y=e^(3t)/(3t), t doesnt = 0. Find the exact value of dy/dx at the point on C where t=2/3 .

To solve this we must use the chain rule which is dy/dt * dt/dx. Firstly, we differentiate dy/dt. For this we must use the quotient rule, this gives us dy/dt=(9te^3t - 3e^3t)/9t^2. Now for dx/dt, by substituting u=2-6t and using the chain rule du/dt * dx/du we get du/dt=-6 and dx/du=-(e^u)/4, we times these together and substitute u=2-6t back in to get dx/dt=(3e^(2-6t))/2. Now we must times dy/dt * 1/(dx/dt) to give us dy/dx=(18te^(3t)-6e^(3t))/(27t^(2)e^(2-6t)). We now must substitute in t=2/3 to give us the exact point on the curve that we require. By doing this and then simplifying our answer we get the value of the gradient dy/dx=(e^4)/2.

LK
Answered by Lauren K. Maths tutor

5757 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrating (e^x)sin(x)


(i) Find the coordinates of the stationary point on the curve y = 3x^2 − 6/x − 2. [5] (ii) Determine whether the stationary point is a maximum point or a minimum point.


A curve has equation -2x^3 - x^2 + 20x . The curve has a stationary point at the point M where x = −2. Find the x-coordinate of the other stationary point of the curve.


Find, using calculus, the x coordinate of the turning point of the curve with equation y=e^3x cos 4


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences