Saturday

August 30, 2014

August 30, 2014

Posted by **Imad** on Thursday, April 8, 2010 at 12:17am.

- calculus -
**Reiny**, Thursday, April 8, 2010 at 12:26amlet's find their intersection points

√2sinx = √2cosx

sinx/cosx = 1

tanx = 1

x = π/4 in the given interval

for y = √2sinx

dy/dx = √2cosx

for x= π/4

dy/dx = √2sinπ/4 = √2(1/√2) = 1

for y = √2cosx

dy/dx = -√2sinx

at x=π/4

dy/dx = -√2(1/√2) = - 1

since the slopes are negative reciprocals of each other when x = π/4, the tangents are perpendicular at that point.

**Related Questions**

Calculus - Find an equation of the tangent line to the given curve at the given ...

Math - Find an equation of the line that bisects the obtuse angles formed by the...

Algebra - How would i complete these problems? 1. (√6mn)^5 2. ^3√16x...

maths - I will be thankfull to you if you help me to solve the below problem: 6...

Algebra Work Check - 1. 5√6 * 1/6√216 5/6 * √1296 5/6 * √...

grade 11 math - I don't get how to do this. I'm so lost please help me. ...

Math - Help with radicals please - Please show work Simplify rational 1. 4√...

Trigonometry - Stuck :( 1. Simplify (5-√-36)(2+√-1) A. 16 − 7i...

algebra 1 - alright i have a few questions if its ok..? Simplify: 1. -2√(...

Pre Calculus 12 - Identify the point on the unite circle coressponding to an ...