Answer:
[tex]\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}[/tex]
Step-by-step explanation:
A road is perpendicular to a highway leading to a farmhouse d miles away.
An automobile passes through the point of intersection with a constant speed [tex]\frac{dx}{dt}[/tex] = r mph
Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.
Then by Pythagoras theorem,
h² = d² + x²
By taking derivative on both the sides of the equation,
[tex](2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}[/tex]
[tex](h)\frac{dh}{dt}=(x)\frac{dx}{dt}[/tex]
[tex](h)\frac{dh}{dt}=rx[/tex]
[tex]\frac{dh}{dt}=\frac{rx}{h}[/tex]
When automobile is 30 miles past the intersection,
For x = 30
[tex]\frac{dh}{dt}=\frac{30r}{h}[/tex]
Since [tex]h=\sqrt{d^{2}+(30)^{2}}[/tex]
Therefore,
[tex]\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}[/tex]
[tex]\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}[/tex]
