Q3 TF. - Calculus

Question

Parametric curves: Imagine the curve traced in the xy-plane by

the coordinates (x, y) = (3z + 1, $z^2$ − 4) as z varies, where the

parameter z is a function of time t.

If the parameter z moves at 5 units per second,

find the instantaneous rate of change of the x- and

y-coordinates as the curve passes through the

point (7, 0).

Step-by-Step Solution

Answer

$\frac{dx}{dt}=15$ units/sec

$\frac{dy}{dt}=20$ units/sec

Step 1. Given Information

$(x,y)=(3z+,z^2-4)$

and $\frac{dz}{dt}=5$ units/sec

Step 2. Calculation

Here, $(x,y)=(3z+1,z^2-4)$

Therefore, $x=3z+1 \& y=z^2-4$

Differentiating the above two equations with respect to time both sides we get,

$\frac{dx}{dt}=3\frac{dz}{dt } \& \frac{dy}{dt}=2z\frac{dz}{dt}$

Since $\frac{dz}{dt}=5 \& (x,y)=(7,0)$

Therefore, $y=0$

or, $z^2-4=0$

Therefore, $z=2$. Since for $z=-2$, $x$ not equals to $7.$

Then $\frac{dx}{dt}=3.5$

$=15$

and $\frac{dy}{dt}=2.2.5$

$=20$