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A curve is defined by the parametric equations 
x(t)=10 cos(9t) and 
y(t)=4e^(10 t). Find 
(dy)/(dx).
Answer:

A curve is defined by the parametric equations x(t)=10cos(9t) x(t)=10 \cos (9 t) and y(t)=4e10t y(t)=4 e^{10 t} . Find dydx \frac{d y}{d x} .\newlineAnswer:

Full solution

Q. A curve is defined by the parametric equations x(t)=10cos(9t) x(t)=10 \cos (9 t) and y(t)=4e10t y(t)=4 e^{10 t} . Find dydx \frac{d y}{d x} .\newlineAnswer:
  1. Find Derivative of x(t)x(t): To find the derivative dydx\frac{dy}{dx} for the parametric equations, we first need to find the derivatives dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} separately.\newlineFor x(t)=10cos(9t)x(t) = 10 \cos(9t), the derivative with respect to tt is dxdt=90sin(9t)\frac{dx}{dt} = -90 \sin(9t).
  2. Find Derivative of y(t)y(t): Now, we find the derivative of y(t)y(t) with respect to tt. For y(t)=4e10ty(t) = 4e^{10t}, the derivative with respect to tt is dydt=40e10t\frac{dy}{dt} = 40e^{10t}.
  3. Calculate (dydx):</b>Thederivative$(dydx)isfoundbydividing$dydt(\frac{dy}{dx}):</b> The derivative \$(\frac{dy}{dx}) is found by dividing \$\frac{dy}{dt} by dxdt\frac{dx}{dt}.dydx=dydtdxdt=40e10t90sin(9t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{40e^{10t}}{-90 \sin(9t)}.
  4. Simplify the Expression: We simplify the expression for dydx\frac{dy}{dx}.dydx=4090×e10tsin(9t)=49×e10tsin(9t)\frac{dy}{dx} = - \frac{40}{90} \times \frac{e^{10t}}{\sin(9t)} = - \frac{4}{9} \times \frac{e^{10t}}{\sin(9t)}.

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