Q 4 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 x-coordinate moves at 5 units per second, find

the instantaneous rate of change of the y-coordinate

as the curve passes through the point (7, 0).

Step-by-Step Solution

Answer

The required instantaneous change in $y-$coordinate is$\frac{dy}{dt}=\frac{20}{3}$

Step 1. Given Information

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

$\frac{dx}{dt}=5$ units/sec

Step 2. Calculation

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

Then $x=3z+1 \& y=z^2-4$

Now, $x=3z+1$

Differentiating both sides with respect to time we get

$\frac{dx}{dt}=3\frac{dz}{dt}$

$or, 5=3\frac{dz}{dt}$

Then, $\frac{dz}{dt}=\frac{5}{3}$

Step 3. Calculation

Since $(x,y)=(7,0)$

Then $y=0$

or, $z^2-4=0$

Therefore, $z=2.$ since $z=-2$ does not gives $x=7.$

From, $y=z^2-4$

or, $\frac{dy}{dt}=2z\frac{dz}{dt}$

or, $\frac{dy}{dt}=2.2.\frac{5}{3}$

Therefore, $\frac{dy}{dt}=\frac{20}{3}$