2018 THREE(b)

Achievement
Question
A curve is defined parametrically by the parametric equations

x=5e2tx = 5e^{2t}

y=2e5ty = 2e^{5t}

Find the gradient of the tangent to this curve at the point where t=0t = 0.
*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
dxdt=10e2t\frac{dx}{dt}=10e^{2t}
dydt=10e5t\frac{dy}{dt}=10e^{5t}
dydx=10e5t10e2t=e5te2t\frac{dy}{dx}=\frac{10e^{5t}}{10e^{2t}}=\frac{e^{5t}}{e^{2t}}
t=0⇒dydx=1t=0\Rightarrow\frac{dy}{dx}=1
Grading Criteria

Achievement (u)

  • Correct solution with correct derivatives.

Merit (r)

-

Excellence T1

-

Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 40:43)
Subscribe
Exam Paper (2018)Assessment Schedule (2018)
Related Questions
2024 THREE(e)Excellence

The diagram below shows part of the graph of the function $f(x) = e^{-x^2}$, where $x \geq 0$. [DI...

2024 THREE(d)Merit

Jamie is doing some baking and pouring the flour to form a conical pile. The height o...

2024 THREE(b)Achievement

The graph below shows the function $y = f(x)$. [DIAGRAM] (i) For the function above, find the valu...

2024 TWO(e)Excellence

The graph of the function $y = \dfrac{xe^{3x}}{2x + k}$, where $k$ is a non-zero constant, has a sin...

2024 TWO(c)Merit

Show that $y = \sin(x^2) - \cos(x)$ is a solution to the equation $\dfrac{d^2y}{dx^2} + 4x^2\,y = 2...