Question

Consider the parametric equations x = eᵗ and y = te^(-t). Find dy/dx and d²y/dx².

a) dy/dx = (e⁻¹ - te⁻¹)/eᵗ and d²y/dx² = d/dx((e⁻¹) - te⁻¹)/eᵗ)
b) dy/dx = (e⁻¹ - te⁻¹)/eᵗ and d²y/dx² = (e^(-t) - te⁻¹)/eᵗ
c) dy/dx = (e⁻¹ - te⁻¹ and d²y/dx² = e⁻¹ - te⁻¹
d) dy/dx = (e⁻¹ - te⁻¹ and d²y/dx² = (e⁻¹ - te⁻¹)/eᵗ

Answer 1

To find dy/dx, differentiate y equation with respect to t and divide it by x equation differentiation with respect to t. dy/dx = (e^(-t) - te^(-t))/e^t. To find d²y/dx² differentiate dy/dx with respect to t. d²y/dx² = (e^(-t) - te^(-t))(-2e^(-t) + 1)

Step-by-step explanation:

To find dy/dx, we need to differentiate the y equation with respect to t and then divide it by the differentiation of the x equation with respect to t. Let's start by finding dy/dt and dx/dt:

dy/dt = d(te^(-t))/dt = e^(-t) - te^(-t)

dx/dt = d(e^t)/dt = e^t

Now, divide dy/dt by dx/dt to find dy/dx:

dy/dx = (e^(-t) - te^(-t))/e^t = (e^(-t) - te^(-t))e^(-t)

To find d²y/dx², we need to differentiate dy/dx with respect to t:

d²y/dx² = d(dy/dx)/dt = d((e^(-t) - te^(-t))e^(-t))/dt

Using the product rule and the chain rule, we can find d²y/dx² = (e^(-t) - te^(-t))(-e^(-t)) + (e^(-t) - te^(-t))e^(-t) = (e^(-t) - te^(-t))(-2e^(-t)) + e^(-t) - te^(-t) = (e^(-t) - te^(-t))(-2e^(-t) + 1)