Question

step 2
we must find dz/dt. differentiating both sides and simplifying gives us the following.
2z 2z·dz/dt = 2x·dx/dt + 2y 2y·dy/dt
dz/dt = 1/z 1/z(x·dx/dt + y y·dy/dt)
step 3
after 3 hours, we have the following.
z = √(x 180² + 75²)
= 5625 195
step 4
therefore, we have the following.
dz/dt = 1/195(180 your answer cannot be understood or graded. more information + 75· your answer cannot be understood or graded. more information)
= mi/h

Explanation:

Step1: Implicit - differentiation

Given an equation (not shown completely here but we assume it's of the form $z^{2}=x^{2}+y^{2}$), differentiating both sides with respect to $t$ using the chain - rule. For the left - hand side, if $F(z)=z^{2}$, then $\frac{dF}{dt}=\frac{d(z^{2})}{dz}\cdot\frac{dz}{dt}=2z\frac{dz}{dt}$. For the right - hand side, $\frac{d(x^{2}+y^{2})}{dt}=\frac{d(x^{2})}{dx}\cdot\frac{dx}{dt}+\frac{d(y^{2})}{dy}\cdot\frac{dy}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$. So, $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$. Then we divide both sides by $2z$ to get $\frac{dz}{dt}=\frac{1}{z}(x\frac{dx}{dt}+y\frac{dy}{dt})$.

Step2: Calculate $z$ value

We know that after 3 hours, if we assume $x$ and $y$ are related to some motion and we have $z = \sqrt{x^{2}+y^{2}}$. Here, $x = 180$ and $y = 75$, so $z=\sqrt{180^{2}+75^{2}}=\sqrt{32400 + 5625}=\sqrt{38025}=195$.

Step3: Substitute values into $\frac{dz}{dt}$ formula

We have $\frac{dz}{dt}=\frac{1}{195}(180\frac{dx}{dt}+75\frac{dy}{dt})$. But we need to know the values of $\frac{dx}{dt}$ and $\frac{dy}{dt}$. Let's assume $\frac{dx}{dt}=v_{x}$ and $\frac{dy}{dt}=v_{y}$. If we assume $\frac{dx}{dt}= 60$ (since if $x$ is distance and in 3 hours it reaches 180, then average speed $v_{x}=\frac{180}{3}=60$) and $\frac{dy}{dt}=25$ (since if $y$ is distance and in 3 hours it reaches 75, then average speed $v_{y}=\frac{75}{3}=25$). Then $\frac{dz}{dt}=\frac{1}{195}(180\times60 + 75\times25)=\frac{1}{195}(10800+1875)=\frac{1}{195}\times12675 = 65$ mi/h.

Answer:

65 mi/h