Rate of decreasing shadow problem solved with derivative and similar triangles

Cars at Intersection Solution
Known values given: x = 4 y = 3 $\frac{dx}{dt}$ = -30 z = $\sqrt{4² + 3²}$ = $\sqrt{25}$ = 5 Use Pythagorean Theorem z² = x² + y² Substitute: y = 3 z² = x² + 3² Take derivative. Solve for $\frac{dz}{dt}$ 2z $\frac{dz}{dt}$ = 2x $\frac{dx}{dt}$ + 0 Substitute: z = 5 x = 4 $\frac{dx}{dt}$ = -30 2 * 5 $\frac{dz}{dt}$ = 2 * 4 * (-30) 10 $\frac{dz}{dt}$ = -240 $\frac{dz}{dt}$ = $\frac{-240}{10}$ = -24 The negative sign indicates the distance is decreasing at 24 mph.
158R

Cars at Intersection
Solution

Known values given:
x = 4 y = 3 $\frac{dx}{dt}$ = -30
z = $\sqrt{4² + 3²}$ = $\sqrt{25}$ = 5

Drawing a parked car and a moving car at an intersection witn x and y coordinates

Use Pythagorean Theorem
z² = x² + y²

Substitute: y = 3
z² = x² + 3²

Take derivative. Solve for $\frac{dz}{dt}$
2z $\frac{dz}{dt}$ = 2x $\frac{dx}{dt}$ + 0

Substitute: z = 5 x = 4 $\frac{dx}{dt}$ = -30
2 * 5 $\frac{dz}{dt}$ = 2 * 4 * (-30)
10 $\frac{dz}{dt}$ = -240
$\frac{dz}{dt}$ = $\frac{-240}{10}$ = -24

The negative sign indicates the
distance is decreasing at 24 mph.

158R

Cars at IntersectionSolution

Cars at Intersection
Solution