# Tofind: how fast the water skiersrises after she leaves the ramp.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 38RE
To determine

Expert Solution

## Answer to Problem 38RE

She rises 7.73ft/s above from the ramp.

### Explanation of Solution

Given:

dhdt=5ft/sec

dxdt=15ft/sec

Concept used:

Slope of the any object is tanθ or dydx .

If sinθ=k then θ=sin-1k or arcsink .

Calculation:

Let

x=the vertical height of the skier.

H=the distance on the ramp (the hypotenuse) from the beginning of the ramp.

Since, dhdt=30ft/s .

The angle of elevation of the ramp is θ=arctan415 .

An equation that relate the quantities is

xh=sin(arctan415) .

x=sin(arctan415)h .

Differentiating with respect to time t

dxdt=sin(arctan415)dhdt .

Plugging the given data:

dxdt=sin(arctan415).30

dxdt=7.73ft/s

Hence, she rises 7.73ft/s above from the ramp.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!