Question

1.4 A curve has parametric equations $x = 2t - \ln(2t)$, $y = t^2 - \ln(t^2)$ where $t > 0$. Find the value of t at the point on the curve where $\frac{dy}{dx} = 2$. (CO2, PO1, C2) [5 Marks]

Name: 14 a curve has parametric equations x 2t ln2t y t2 lnt2 where t 0 find the value of t at the point on the curve where fracdydx 2 co2 po1 c2 5 marks 09646
Uploaded: 2024-04-19T16:23:25-08:00
Duration: 3 min 24 s
Channel: Audrey Fong
Description: 14 a curve has parametric equations x 2t ln2t y t2 lnt2 where t 0 find the value of t at the point on the curve where fracdydx 2 co2 po1 c2 5 marks 09646

          1.4 A curve has parametric equations $x = 2t - \ln(2t)$, $y = t^2 - \ln(t^2)$ where $t > 0$. Find the value of t at the point on the curve where $\frac{dy}{dx} = 2$. (CO2, PO1, C2)
[5 Marks]

$14 a curve has parametric equations x 2t ln2t y t2 lnt2 where t 0 find the value of t at the point on the curve where fracdydx 2 co2 po1 c2 5 marks 09646$

Added by Vicenta C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Audrey Fong

04/19/2024

Step 1

We can use the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, find $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{d}{dt}(2t - \ln(2t))$ $\frac{dx}{dt} = 2 - \frac{1}{2t} \cdot 2$ $\frac{dx}{dt} = 2 - \frac{1}{t}$ Next, find $\frac{dy}{dt}$: $\frac{dy}{dt} = Show more…

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1.4 A curve has parametric equations $x = 2t - \ln(2t)$, $y = t^2 - \ln(t^2)$ where $t > 0$. Find the value of t at the point on the curve where $\frac{dy}{dx} = 2$. (CO2, PO1, C2) [5 Marks]