a. Find dydx when x = 2at and y = 2at^2 b. Evaluate: 2x 3x \, dx c. Determine the interval in which the function f(x) = 12x^2 x is increasing or decreasing. GROUP: B Short answer questions 5 × 2 × 4…

a. Find $\frac{dy}{dx}$ when $x=2at$ and $y=2a{t}^2$

We are given: $$x=2at,y=2a{t}^2$$

To find $\frac{dy}{dx}$, we use the chain rule for parametric differentiation: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

First, compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $$\frac{dx}{dt}=\frac{d}{dt}(2at)=2a$$ $$\frac{dy}{dt}=\frac{d}{dt}(2a{t}^2)=4at$$

Now, substitute these into the chain rule formula: $$\frac{dy}{dx}=\frac{4at}{2a}=2t$$

Final answer: $\frac{dy}{dx}=\boxed{2t}$

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b. Evaluate: $\int \sin 2x\cdot \sin 3xdx$

We use the trigonometric identity for the product of sines: $$\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$$

Let $A=2x$ and $B=3x$. Then: $$\sin 2x\sin 3x=\frac{1}{2}[\cos (2x-3x)-\cos (2x+3x)]=\frac{1}{2}[\cos (-x)-\cos (5x)]$$

Since $\cos (-x)=\cos x$, we have: $$\sin 2x\sin 3x=\frac{1}{2}[\cos x-\cos 5x]$$

Now, integrate term by term: $$\int \sin 2x\sin 3xdx=\frac{1}{2}\int (\cos x-\cos 5x)dx$$ $$=\frac{1}{2}\left[\int \cos xdx-\int \cos 5xdx\right]$$ $$=\frac{1}{2}\left[\sin x-\frac{1}{5}\sin 5x\right]+C$$ $$=\frac{1}{2}\sin x-\frac{1}{10}\sin 5x+C$$

Final answer: $\boxed{\frac{1}{2}\sin x-\frac{1}{10}\sin 5x+C}$

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c. Determine the interval in which the function $f(x)=\frac{1}{2}{x}^2-x$ is increasing or decreasing.

To determine where the function is increasing or decreasing, we find its derivative and analyze its sign.

First, compute the derivative: $$f(x)=\frac{1}{2}{x}^2-x$$ $$f^{\prime }(x)=\frac{d}{dx}\left(\frac{1}{2}{x}^2-x\right)=x-1$$

Now, find the critical point by setting $f^{\prime }(x)=0$: $$x-1=0⟹x=1$$

We analyze the sign of $f^{\prime }(x)$ in the intervals determined by the critical point:

Final answer:

• The function $f(x)$ is decreasing on the interval $\boxed{(-\infty ,1)}$.
• The function $f(x)$ is increasing on the interval $\boxed{(1,\infty )}$.