Question

suppose $y = \sqrt{2x + 1}$, where $x$ and $y$ are functions of $t$. (a) if $\frac{dx}{dt}=15$, find $\frac{dy}{dt}$ when $x = 4$. $\frac{dy}{dt}=$ (b) if $\frac{dy}{dt}=4$, find $\frac{dx}{dt}$ when $x = 12$. $\frac{dx}{dt}=$ 5. -/1 points a plane flying horizontally at an altitude of 1 mile and a speed of 540 mi/h passes directly over a radar station. find the rate at which the distance from the plane to the station is increasing when it has a total distance of 5 miles away from the station. (round your answer to the nearest whole number.) mi/h

Explanation:

Step1: Differentiate y with respect to t

Using the chain - rule, if $y=\sqrt{2x + 1}=(2x + 1)^{\frac{1}{2}}$, then $\frac{dy}{dt}=\frac{1}{2}(2x + 1)^{-\frac{1}{2}}\cdot2\cdot\frac{dx}{dt}=\frac{1}{\sqrt{2x + 1}}\cdot\frac{dx}{dt}$.

Step2: Solve part (a)

Given $\frac{dx}{dt}=15$ and $x = 4$. Substitute into the derivative formula: $\frac{dy}{dt}=\frac{1}{\sqrt{2\times4+1}}\times15=\frac{15}{\sqrt{9}}=\frac{15}{3}=5$.

Step3: Rearrange the derivative formula for part (b)

From $\frac{dy}{dt}=\frac{1}{\sqrt{2x + 1}}\cdot\frac{dx}{dt}$, we can solve for $\frac{dx}{dt}$: $\frac{dx}{dt}=\sqrt{2x + 1}\cdot\frac{dy}{dt}$.

Step4: Solve part (b)

Given $\frac{dy}{dt}=4$ and $x = 12$. Substitute into the formula for $\frac{dx}{dt}$: $\frac{dx}{dt}=\sqrt{2\times12 + 1}\times4=\sqrt{25}\times4=5\times4 = 20$.

Answer:

(a) 5
(b) 20