Question

assume that x=x(t) and y=y(t). let y=x2+3 and dx/dt=4 when x=5.find dy/dt when x=5

Answer 1

SOLUTION

GIVEN

Assume that x = x(t) and y = y(t)

Let y = x² + 3

$\displaystyle \sf \: \frac{dx}{dt} = 4 \: \: \: when \: \: x = 5$

TO DETERMINE

$\displaystyle \sf \: \frac{dy}{dt} \: \: \: when \: \: x = 5$

EVALUATION

Here it is given that

y = x² + 3 - - - - - - (1)

Also given that

$\displaystyle \sf \: \frac{dx}{dt} = 4 \: \: \: when \: \: x = 5 \: \: \: \: \:$

Differentiating both sides of Equation 1 with respect to t we get

$\displaystyle \sf \: \frac{dy}{dt} = 2x \frac{dx}{dt} + 0$

Putting x = 5 we get

$\displaystyle \sf \implies \frac{dy}{dt} = 2 \times 5 \times 4$

$\displaystyle \sf \implies \frac{dy}{dt} = 40$

FINAL ANSWER

$\displaystyle \sf \frac{dy}{dt} = 40 \: \: \: \: when \: \: x = 5$

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