consider the curve parametrized by $x = 1+ln(t)$ and $y=t^{4}+t$.\na) first eliminate the parameter $t$ to…

consider the curve parametrized by $x = 1+ln(t)$ and $y=t^{4}+t$.\na) first eliminate the parameter $t$ to compute $\frac{dy}{dx}$ in terms of $x$ and $y$.\nb) next, use the fact that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ to compute $\frac{dy}{dx}$ in terms of $t$.\nc) lastly, use either your answer from part a) or part b) to find the slope of the tangent line when $t = 1$ and $(x,y)=(1,2)$.

consider the curve parametrized by $x = 1+ln(t)$ and $y=t^{4}+t$.\na) first eliminate the parameter $t$ to compute $\frac{dy}{dx}$ in terms of $x$ and $y$.\nb) next, use the fact that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ to compute $\frac{dy}{dx}$ in terms of $t$.\nc) lastly, use either your answer from part a) or part b) to find the slope of the tangent line when $t = 1$ and $(x,y)=(1,2)$.

Answer

Explanation:

Step1: Solve for (t) from (x) - equation

Given (x = 1+\ln(t)), we can rewrite it as (\ln(t)=x - 1), then (t = e^{x - 1}). Substitute (t) into (y) - equation: (y=(e^{x - 1})^{4}+e^{x - 1}=e^{4(x - 1)}+e^{x - 1}). Now, differentiate (y) with respect to (x) using the chain - rule. (\frac{dy}{dx}=4e^{4(x - 1)}+e^{x - 1}). But we can also express it in terms of (x) and (y) in a non - simplified way. First, from (y=t^{4}+t) and (t = e^{x - 1}), we have (\frac{dy}{dx}=\frac{4t^{3}+1}{\frac{1}{t}}=t(4t^{3}+1)=4t^{4}+t). Since (y=t^{4}+t) and (t^{4}=y - t), and (t = e^{x - 1}), (\frac{dy}{dx}=4(y - t)+t=4y-3t=4y - 3e^{x - 1}).

Step2: Differentiate (x) and (y) with respect to (t)

Differentiate (x = 1+\ln(t)) with respect to (t): (\frac{dx}{dt}=\frac{1}{t}). Differentiate (y=t^{4}+t) with respect to (t): (\frac{dy}{dt}=4t^{3}+1). Then, by the formula (\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}), we have (\frac{dy}{dx}=t(4t^{3}+1)=4t^{4}+t).

Step3: Find the slope at (t = 1)

Using the result from part b) (it's easier in this case), when (t = 1), substitute (t = 1) into (\frac{dy}{dx}=4t^{4}+t). Then (\frac{dy}{dx}\big|_{t = 1}=4\times1^{4}+1=5).

Answer:

a) (\frac{dy}{dx}=4y-3e^{x - 1}) b) (\frac{dy}{dx}=4t^{4}+t) c) (5)