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In Exercises 41-44 find an equation for the line tangent to the parametric curve at the given value ot f.

x=cos3t,y=sin3t,t=π4.

Short Answer

Expert verified

The slope of the parametric equation is dydx=m=-3.

Step by step solution

01

Given information

The parametric curve is x=cos3t,y=sin3t,t=π4.

02

Calculation

Consider the parametric curves x=t2,y=(2-t)2at t=12.

The objective is to find the equation of a tangent line for the given parametric equations.

The formula to find the tangent line equation is y-y1=mx-x1.

First find the slope of the parametric curves by finding the derivative of the parametric curves.

For that we use the formuladydx=dydtdxdt.

Now take the parametric equation x=t2.

Differentiate the curve with respect to t.

Then

dxdt=ddtt2since x=t2

dxdt=2t·dtdtsincedt2dt=2t

dxdt=2t

03

Further simplification

Now take the parametric equation y=(2-t)2.

Differentiate the curve with respect to t.

dydt=ddt(2-t)2dydt=2×(2-t)ddt(2-t)dydt=2(2-t)(-1)dydt=-2(2-t)

Now substitute the values of dxdt,dydtin the slope formula dydx=dydtdxdt. Then

dydx=-2(2-t)2tSincedxdt=2t,dydt=-2(2-t)

dydx=-(2-t)t

The slope when t=12is as follows,

dydx1=12=-2-1212dydx1=12=-4-1212

On further simplification,

dydx1-12=-312dydx1,12=-3

Thus, the slope of the parametric equation is dydx=m=-3.

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