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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

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Section
BuyFindarrow_forward

Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

Finding Slope and Concavity In Exercises 39 - 46. find dy/dx and d 2 y / d x 2

, and find the slope and concavity (if possible) at the given value of the parameter.

Parametric Equations Parameter

x = 1 + 6 t , y = 4 5 t t = 3

To determine

To calculate: The value of dydx,d2ydx2, slope and concavity for the parametric equation x=2+5t,y=14t at t=3.

Explanation

Given:

The parametric equation is x=2+5t,y=14t and the condition is at t=3.

Formula used:

Parametric form of first derivative is,

dydx=(dydt)(dxdt)

Parametric form of second derivative is,

d2ydx2=ddt(dydx)dxdt

For concavity, if d2ydx2>0, then graph is concave upward and if d2ydx2<0, then graph is concave downward.

Calculation:

Consider the parametric equation x=2+5t,y=14t and the condition t=3.

Now for the value of dydx, divide numerator and denominator by dt as both x and y are functions of t,

dydx=dydtdxdt

Differentiate x=2+5t and y=14t with respect to t and put them in below equation.

dydx=(dydt)(dxdt)=040+5=45

Since, the parameter t is missing from dydx.

So at any value of t, dydx is 45.

Also dydx represent the slope for any graph, so the slope is 45.

Now for the value of d2ydx2, again differentiate dydx with respect to t.

d2ydx2=(ddt(dydx))(dxdt)=ddt(45)5=05=0

Thus, the value of d2ydx2 is 0.

Recall that, for concavity, if d2ydx2>0, then graph is concave upward and if d2ydx2<0, then graph is concave downward.

Since, the value of d2ydx2 is 0 so the graph of parametric equation x=2+5t,y=14t at t=3 is neither concave upward nor concave downward.

From the equations x=2+5t,y=14t.

Find the value of y in terms of x;

y=14ty1=4t

Solve further,

y1(4)=t

Put t in equation x=2+5t, to get;

x=2+5(y1(4))4x=8+5y54x5y=13

This gives;

4x+5y=13

Thus, the parametric equation gives the equation of line

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