Megan cycles from her home, \(H\), to school, \(S\), each day.
She rides along a path from her home to point \(P\) at a constant speed of \(10\) kilometres per hour. At point \(P\), Megan cuts across a park, heading directly to school. When cycling across the park, Megan can only cycle at \(6\) kilometres per hour.
The path distance \(HQ\) is \(4\) km, and the vertical distance \(QS\) is \(2\) km.
At what distance from her home should she choose to cut across the park in order to make her travelling time a minimum?
You must use calculus and show any derivatives that you need when solving this problem.
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
Step 1
Choose a variable
Let \(x\) be the distance from \(H\) to \(P\) along the road. What is the remaining horizontal distance from \(P\) to \(Q\)?
Step 2
Find the shortcut distance
Use Pythagoras on the triangle \(PQS\).
Preview will appear here.
Step 3
Write the total time function
Add the time on the road and the time across the park. Time is distance divided by speed.
Preview will appear here.
Step 4
Differentiate carefully
The square-root term needs the chain rule, so it helps to rewrite it as a power first. Choose the correct simplified derivative.